rt 


f 


^  H  "^o^' 


IN  MEMORIAM 
FLORIAN  CAJORl 


fU^ 


I 


American  Progressive   Series, 

THE 

COMPLETE 


ARITHMETIC 


BY 

MILTON    B.    GOFF,    A.M., 

M 

PROFESSOR  OF  MATHEMATICS   IN   THE   WESTERN   UNIVERSITY   OF   PENNSYLVANIA- 


PITTSBURGH  : 

PUBLISHED  BY  H.  I.  GOURLEY,  Agt 

98    FOURTH    AVENUE. 


Copyright,  1876.     A.  H.  ENGLISH  &  CO. 


LANE  S.   HART, 

PRINTER  AND  BINDER, 

HARRISBURG,  PA. 


CAJORI 


BY  an  experience  of  over  twenty  years  in  the  class-room,  the  author  has  been 
convinced  of  these  three  things : 

1.  That  both  in  our  public  schools  and  academies  too  many  books  hav2  been  used  - 
and  too  much  time  has  been  spent  in  the  study  of  Arithmetic. 

2.  That  while  the  study  of  Intellectual  Arithmetic,  as  such,  is  doubtless  benefi- 
cial, the  greatest  good  results  from  its  study  in  close  connection  with  Written  Arith- 
metic ;  indeed,  that  the  two  cannot  be  separated. 

3.  That  while  it  is  impossible  from  any  text-book,  to  teach  a  pupil  all  '*  that  he  is 
to  practice  as  a  man,"  it  is  possible  to  comprise  in  less  space  than  is  usua"y  done, 
and  in  a  manner  that  will  serve  to  strengthen  the  reasoning  powers,  many  of  the 
facts  with  which  our  youth  will  have  to  deal  in  after  life. 

The  aim,  therefore,  in  the  following  pages  has  been  to  present  in  a  form  compact, 
but  not  obscure,  all  that  is  necessary  for  the  instruction  of  youth  in  the  science  of 
Arithmetic,  whether  as  a  preparation  for  the  ordinary  vocations  of  life,  or  as  a  pre- 
liminary training  for  a  course  in  Mathematics. 

The  First  Book  in  Arithmetic  is  intended  for  the  younger  pupils,  and  may  be 
placed  in  their  hands  at  a  very  early  stage  of  advancement  in  their  education. 
And  it  is  safe  to  say,  that  those  finishing  the  First  Book  will  readily  mastc"  the 
second,  or  Complete  Arithmetic.  It  is  believed,  however,  that  in  the  rural  dis- 
tricts, as  children  do  not  enter  school  at  as  tender  an  age  as  in  the  cities,  the  present 
work  will  be  found  sufficiently  elementary  in  the  first  part,  even  for  beginners  ;  and 
although  in  the  body  of  the  work  the  matter  is  often  as  difficult,  and  quite  as  exten- 
sive, as  in  the  Higher  Arithmetic,  the  whole  is  arranged  in  so  progressive  a  manner, 
that  the  natural  and  necessary  development  of  the  mind  of  the  pupil  will  enabl© 
him  to  master  every  difficulty. 

We  cannot  see  why  Oral  Arithmetic  and  Written  Arithmetic  should  be  sepa- 
rated ;  for  every  process  in  the  latter  includes,  more  or  less,  the  former.  Yielding, 
therefore,  to  what  seemed  to  be  a  necessity,  we  have  made  the  "  Oral "  and  "  Writ- 
ten" alternate  in  every  case  where  it  was  deemed  judicious,  presenting  as  great  a 
variety  of  examples  as  is  usually  found  in  works  devoted  exclusively  to  the  Intel- 
lectual, and  better  adapted,  in  our  opinion,  for  teaching  the  pupil  to  think.  At  the 
same  time,  the  "  Oral"  part  can,  if  desired,  be  omitted  without  destroying  connec- 
tion between  subjects. 

The  third  object  has  been  kept  steadily  in  view.  Examples  and  problems  have 
been  chosen  with  special  reference  to  the  wants  of  our  American  youth.  And 
while  we  do  not  claim  perfection  in  this  matter,  we  do  think  we  have  taken  a  step  in 
the  right  direction.  We  believe  that  no  man  whose  son  is  to  engage  in  any  mechan- 
ical pursuit,  will  begrudge  the  space  (249-257)  given  to  the  computation  of 
carpenters',  bricklayers',  painters',  and  stonemasons'  work.  Nor  will  any  one  hav- 
ing to  do  with  lands  in  the  Western  States,  regard  the  time  devoted  to  Government 

iii 


918305 


IV  PREFACE. 

Lands  (21 1^  248)  as  time  poorly  spent.  These  articles  we  regard  as  being  "  for 
the  greatest  good  of  the  greatest  number";  since  their  position,  somewhat  early  in 
the  work,  serves  the  purpose  of  giving  to  many  who  would  never  see  them  if  placed 
on  the  last  pages  with  Mensuration,  an  idea  of  what  they  shall  have  to  labor  at  in 
after  years.    Besides,  the  rules  may  be  laid  by  for  future  use  and  reference. 

In  addition  to  the  points  named,  we  wish  to  call  attention  to  our  Outlines.  It 
will  be  observed  that  we  have  a  General  Outline  and  for  each  chapter  a  Sub-outline. 
And  although  these  may  not  be  all  that  the  heart  could  desire,  we  do  think  that  every 
teacher  may  make  them  a  valuable  aid  in  the  prosecution  of  his  labors.  We  like 
them  better  than  synopses  placed  at  the  close  of  the  chapter  ;  for  the  mind,  like  the 
eye,  takes  in  an  object  as  a  whole,  then  separates  it  into  parts,  then  sub-divides  each 
part,  and  so  proceeds,  until  all  the  minutiae  are,  in  turn,  examined.  This  is  the  prin- 
cipal object  of  the  Outlines  ;  though  they  also  serve  the  purpose  of  a  ready  index. 

In  the  arrangements  of  subjects,  deviation  has  been  made  from  the  "  beaten 
track";  but  only  with  great  caution.  The  placing  of  Decimals  immediately  after 
Division  of  Integers  was  decided  upon  only  after  careful  experiment  by  many  of 
our  most  experienced  educators. 

And  last,  though,  perhaps,  to  some,  not  least,  is  the  fact  that  although  the  publish- 
ers have  spared  neither  pains  nor  expense  (as  the  most  casual  examination  of  the 
books  themselves  shows)  to  make  this  series,  consisting  of  the  First  Book  in  Arith- 
metic and  The  Complete  Arithmetic,  second  to  none,  they  propose  to  offer  to 
their  customers  in  these  two  volumes,  what  is  usually  found  in  from  three  to  five, 
and  at  correspondingly  low  prices ;  and  thus  relieve  our  ''  humble  poor,"  as  well  as 
those  of  moderate  means,  of  much  of  the  expense  they  can  so  ill  afford  in  the  procur- 
ing of  text-books  for  their  children.  They  do  not  expect,  however,  to  change  in  a 
day  the  habits  strengthened  by  years.  There  will  be  men  who  will  differ  from 
them  and  us  in  opinion,  and  who  will  think,  despite  what  has  been  said,  that  two 
books  are  not  enough ;  that  Intellectual  and  Written  Arithmetic  should  be  sepa^- 
rate  works.  With  these  the  publishers  have  no  quarrel,  but,  having  acted  in  accord- 
ance with  their  best  judgment,  now  submit  to  an  intelligent  public  for  their  decision 
this  question  so  important  to  all  interested  in  the  subject  of  education. 

One  word  to  our  fellow-teachers.  We  have  made  an  earnest  effort  to  present  such 
a  work  as  will  meet  with  your  approval  and  suit  your  wants.  From  comments  of  some 
of  you  on  portions  of  the  manuscript  and  proof-sheets  that  you  have  seen,  we  have 
reason  to  believe  that  you  will  be  pleased  with  the  result  of  our  labors.  But  we 
cannot,  and  do  not,  expect  you  all  to  agree  with  us  in  all  points.  You  have  minds 
of  your  own ;  and  we  are  glad  to  know  that  you  have  not  only  the  independence 
to  think  for  yourselves,  but  that  you  possess  that  liberality  of  \iew,  that  grants 
others  the  privileges  you  claim  for  yourselves.  Therefore,  while  we  do  not  hope 
to  escape  your  criticism,  we  do  look  for  that  honest,  straightforward  expression 
of  opinion  that  becomes  those  who  are  engaged  in  a  profession  which,  when  prop- 
erly pursued,  develops  the  noblest  qualities  of  mind  and  heart. 

In  the  preparation  of  this  work,  we  have  been  greatly  aided  and  encouraged  by 
many  friends,  whose  kind  suggestions  have  been  thankfully  received  and  freely 
used.  To  all  we  return  our  best  thanks,  as  well  as  to  our  co-laborers,  Messrs.  J.  M. 
Logan  and  H.I.  Gourley,  of  the  Pittsburgh  Schools,  to  whose  superior  taste  we 
are  indebted  for  the  neat  arrangement  of  headings,  plates,  and  outlines,  and  by 
whose  care  and  vigilance  many  errors  and  crudities  have  been  avoided. 

Wbstkrw  University  of  Pa.,  June^  1876.  M.  B.  G. 


''-F^^, 


PAGE 

Foggestions 7 

CHAPTER     I. 

INTEGKRS. 

O  UTI.IJVE  of  Arithmetic 11 

General  Definitions 12 

OUTLIA^E  of  Numeration  and 

Notation 14 

Arabic  Numeration  and  Notation.    .  15 

Roman  Numeration  and  Notation  ...  21 

<9  ^7J?'ZyJK^  of  Addition 23 

Addition 24 

O  UTLIJVE  of  Subtraction 87 

Subtraction 38 

O  UTLIJ\rE  of  Multiplication  ....  47 

Multiplication 48 

Review  Problems 59 

O  UrZIJVE  of  Division 60 

Division 61 

Problems  Under  the  Four  Rules ...  73 

CHAPTER    II. 

DECIMALS. 

OUTm^E 79 

Numeration  and  Notation 80 

Addition 87 

Subtraction 90 

Multiplication  92 

Division 95 


FAGB 

United  States  Monet 101 

Addition  103 

Subtraction 105 

Multiplication 107 

Division 109 

CHAPTER    III. 

FRACTIONS. 

OUTLINE Ill 

Numeration  and  Notation , .  112 

Reductions 117 

Greatest  Common  DiviBor 119 

Least  Common  Multiple 121 

Cancellation 124 

Addition 138 

Subtraction , 142 

Multiplication 145 

Division 151 

Review  Problems 157 

Converse  Reductions 162 

Aliquot  Parts 164 

Bills 167 

CHAPTER       IV. 

DENOMINATE    NUMBERS. 

OUTLIJ^E 171 

Definitions 173 

Tables— Moneys 175 

Weights  180 

Measures ....  186 


yi 


CONTENTS. 


PAOK 

Reductions 202 

Addition 223 

Subtraction 230 

Multiplication 238 

DiviBion 241 

APPLICATIONS  OF  MEAS- 
URES. 

O  UTLIJVB 244 

Square  Measure 245 

Artificers' Work 249 

Cubic  Measure 254 

E:xcavations  and  Embankments.  254 

Masonry— Stone-work 255 

Brick-work 257 

Capacities 259 

Time 262 

CHAPTER    V. 

RATIO    AND    PROPORTION. 

OUTLIJVB 267 

Ratio 268 

Variation  271 

Proportion— Simple 274 

Compound 280 

Distributive 284 

CHAPTER   VI. 

PERCENTAGE. 

OUTLIJ^S 289,  290 

Definitions 291 

General  Cases 293 

APPLICATIONS. 

Profit  and  Loss 303 

Insurance 315 

Commission 318 

Simple  Interest 322 

Partial  Payments 339 

Compound  Interest 343 


PAGE 

Discount— Bank 349 

True 354 

Stocks 357 

Exchange 362 

Taxes  and  Duties 367 

Direct , 368 

Indirect 369 

Partnership 371 

Bankruptcy 375 

Average  of  Payments 376 ' 

CHAPTER    VII. 

INVOLUTION   AND  EVOLU- 
TION. 

OVTLIJV^ 383 

Involution 384 

Evolution 386 

Square  Koot 387 

Applications 390 

Cube  Root 893 

Applications" 399 

CHAPTER    VIII. 

PROGRESSIONS. 

OUTLIJ\rJE 400 

Arithmetical 401 

Geometrical 404 

Annuities 406 

CHAPTER    IX. 

MENSURATION. 

OUTLIJ\rE 409 

Square  Measure 410 

Solids 415 

Surfaces  of  Solids 417 

Volumes  of  Solids 421 

METRIC  SYSTEM 42C 

LUMBERMEN'S  RULES 42g 

MISCELLANEOUS  PROBLEMS 42( 


Art.  1-19.  The  teacher  must  exercise  discretion  in  use  of  definitions. 
Those  in  Section  T.  need  not  all  be  committed  at  once,  as  some  of  them  are  given  in 
the  body  of  the  work  as  needed. 

55.  Addends^  although  a  comparatively  new  term,  is  not  used  without 
authority. 

61 .  In  Mental  Exercises,  use  the  model  best  adapted  to  each  pupil.  If  deemed 
best,  give  the  younger  pupils  the  mental  problems  as  dictation  exercises  on  the 
slate. 

73.     Refer  to  definitions  (4  and  5),  or  explain  fully  concrete  and  abstract. 

78.  As  early  as  possible,  it  is  well  to  show  that  the  placing  of  the  subtrahend 
under  the  minuend,  is  a  matter  of  convenience. 

89.  When  multiplying  we  may  regard  both  terms  as  abstract,  and  then  attach 
to  the  product  such  name  as  the  nature  of  the  question  demands.  Thus,  since 
1  bbl.  flour  costs  $8,  9  bbls.  cost  9  times  as  much,  or  $72.  9x8  =  8x9=  72.  The 
answer  must  be  dollars.    Therefore,  $72,  answer. 

93.    Prob.  10  may  be  contracted  thus : 


374781 
1402 

749562 


525442962 


First  multiply  by  2  ;  then  multiply  this  product  by  7,  placing  the 
right-hand  figure  under  4,  and  the  remaining  figures  in  order  to  the 
left ;  for  2  X  7  =  14.  This  is  given  merely  as  a  sample.  Once  show 
the  pupil  how  to  lighten  his  labor,  and  generally  he  is  not  slow  to  take 
advantage  of  any  short  process ;  and  will  soon  reap  great  benefit. 
Try  Prob.  21. 


Many  operations  in  multiplication  may  be  solved  by  such  devices  as  this: 


324  X  81=? 


26244 


324  X  18=? 


324  X  108=? 
2592 


324  X  801  =  ? 


95.  When  ciphers  are  on  the  right  of  significant  figures,  either  in  the  multipli- 
cand or  multiplier,  or  both,  they  should  not  be  considered  until  the  significant  fig 
ures  are  multiplied  together,  when  the  ciphers  must  be  annexed  to  the  product 

rii 


Vlll  SUGGESTIONS. 

102.  Example  Second.  This  may  also  be  solved  thus:  If  each  of  two  per- 
sons receive  $1,  to  divide  $10  equally  between  them,  each  must  receive  as  many 
times  $1  as  $2  is  contained  times  in  $10,  or  5  times  $1  =  $5.  This  relieves  us  of  the 
inconsistency  of  calling  the  divisor  an  abstract  and  the  dividend  a  coticrete 
number. 

Or,  since  taking  one-half  oi  %Vi  is  the  same  as  multiplying  $10  by  J,  we  have  the 
multiplicand  and  product  of  the  same  kind  (89). 

106-108.  Show  that  the  divisor  may  be  written  in  any  other  convenient 
place  as  well  as  at  the  left  or  right  of  the  dividend. 

Problems.  The  earnest  teacher  will  not  fail  to  supply  the  student  with  abun- 
dant examples.  Pp.  73-78  are  thought  to  afford  a  fair  variety  of  such  examples  as 
the  pupil  needs  to  make  him  thoroughly  familiar  with  the  principles  and  operations 
already  discussed. 

112-119.  Show  the  strong  resemblance  of  a  whole  number,  or  Integer,  and  a 
decimal ;  and  exhibit  in  the  strongest  light  the  importance  of  the  decimal  point. 

120.  The  Second  Method  of  Numeration  and  Notation  possesses  such  great 
advantages  over  the  first  method  that  we  wonder  at  the  limited  use  of  the  former. 

129-1 38 .  Too  much  care  cannot  be  exercised  in  teaching  "  Multiplication  of 
Decimals  "  and  "  Division  of  Decimals  ";  and  no  part  of  the  Arithmetic  will  better 
repay  this  care ;  for,  pupils  once  thorough  in  these,  will  move  along  easily  and 
rapidly. 

138.  Reducing  dividend  and  di%'isor  to  the  same  denomination  before  divid- 
ing has  the  advantage  of  clearness,  but  is.  sometimes  inconvenient  in  practice. 
Thus  :  Divide  1728  by  1.2.  Annexing  .0  to  1728,  we  have  1728.0,  dividend  and  divi- 
sor, both  tenths^  and  the  quotient  a  whole  number,  1440.  Dividing  .0001728  by  1.2 
by  this  method,  however  plain  it  may  be,  is  a  clumsy  performance. 

167.  In  finding  the  factors  of  a  number,  ^&  facts  in  this  article  may  be  greatly 
extended,  as  for  example  :  Fourth.  Any  even  number,  the  sum  of  whose  digits  is 
divisible  by  3,  has  6  for  an  exact  divisor  ;  Fifth,  Any  number,  whose  two  right- 
hand  figures  are  divisible  by  4,  has  4:  for  an  exact  divisor,  etc.,  etc. 

183.  Special  Rules  on  p.  148  are  given,  that  teacher  and  pupil  may  have  both 
variety  and  choice.    We  prefer,  however,  the  General  Rule. 

184.  The  same  remarks  apply  to  Rules  on  pp.  154,  155. 

185.  The  pupil  should  be  able  at  once  to  change  a  decimal  into  a  common  frac- 
tion, or  a  common  fraction  into  a  decimal. 

186.  187»  are  eminently  practical  and  should  be  thoroughly  mastered. 

192-232  embrace  the  tables  used  in  Denominate  Numbers,  and  are  placed 
together  as  a  matter  of  convenience  for  easy  reference.  The  exercises  which  follow 
are  placed  under  their  appropriate  headings,  so  that  a  table,  or  a  convenient  number 
of  tables,  with  the  exercises  Belonging  to  each,  may  be  readily  assigned  as  a  lesson. 

193.  The  Table  of  Federal  Money,  together  with  definitions,  are  here  in- 
serted to  preserve  the  uniformity  of  the  system. 

The  representations  of  the  coins  of  the  United  States  are  all  that  are  coined  at 
the  present  time.    Of  the  coins  of  Great  Britain,  Germany,  and  France,  only  a  lim- 


SUGGESTION'S.  ix 


ited  number  are  inserted.    It  is  advisable,  when  possible,  for  the  teacher  to  exhibit 
the  actual  coins.    The  same  is  true  of  all  the  weights  and  measures  represented. 

210,  211,  216.  Show  how  square  and  cubic  measure  stand  related  to  Long 
measure. 

211.  We  see  no  good  reason  why  a  pupil  should  wait  until  he  studies  a  treat- 
ise on  surveying  before  he  knows  how  a  township  is  subdivided. 

By  means  of  cut,  p.  192,  the  teacher  may  suggest,  and  the  pupil  solve,  a  great 
number  of  interesting  problems.  By  means  of  Section  Maps  of  the  Western  States, 
the  pupil  may  locate  the  principal  cities,  towns,  etc. 

214.  The  edge  of  a  cube  is  one  of  its  dimensions.  The  edge  of  any  solid  is  a 
line  formed  by  the  meeting  of  two  adjacent  faces  of  that  solid. 

245.  Although  three  methods  are  given,  whether  the  pupil  shall  study  them 
all  at  once  should  be  determined  by  his  stage  of  advancement. 

246,  257,  are  entirely  practical,  and  while  intended  for  pupils  in  general,  are 
especially  useful  for  those  who  have  no  prospect  of  taking  a  course  in  Mathematics, 
or  who  cannot  even  find  the  time  to  go  as  far  as  Mensuration  proper,  as  treated  in 
the  last  pages  of  this  work.  Much  is  given  here  in  a  convenieni  form  that  is  not 
easy  to  find  in  any  one  book. 

248.  Nearly  every  pupil  in  the  Western  States  will  be  interested  in  knowing 
the  manner  of  dividing  lands  m  his  own  county. 

257.  Prob.  4.    Smce  when  the  sides  were  15  inches  high,  the  wagon-bed  held 

50 
43^1  cubic  feet ;  in  order  to  hold  50  cubic  feet,  it  must  be  -^^  of  15  inches  high,  or 

17|t  inches.    Hence  the  depth  is  increased  2|?  in. 

Probs.  13, 17  and  18  may  be  solved  in  the  same  manner. 

258.  This  may  be  illustrated  by  the  change  apparent  in  a  wptch,  in  going 
from  San  Francisco  to  New  York,  New  York  to  Liverpool,  Liverpool  to  Canton 
(China),  Canton  to  San  Francisco  ;  illustrating  that  a  good  time-keeper  will  lose  a 
day  in  going  round  the  world  eastwardly  ;  and,  in  like  manner,  going  westwardly 
will  gain  a  day. 

259.  The  word  measure  is  here  used  in  its  general  sense;  for  although  25 
consists  of  a  number  of  parts,  yet,  for  the  purpose  of  measuring,  it  is  a  unit.  Fof 
example  :  How  many  centals  in  2050  lbs.  of  wheat  ?  Here  we  divide  2050  by  100, 
the  number  of  lbs.  in  one  cental,  and  100  lbs.  is  regarded  as  the  measuring  unit. 

286.  P.  285,  Ex.  Equimultiples  of  numbers  are  the  products  of  those  num- 
bers by  a  given  number.    Thus  :  7  x  5  and  8x5  are  equimultiples  of  7  and  8. 

291,  292.     Note  especially  the  difference  between  Rate  per  cent,  and  Rate. 

305.  In  Higher  Mathematics  and  Applications,  Formulas  are  deemed  invalu- 
able.   Why  not  in  Arithmetic  ? 

319.  Solution  of  Ex.  1,  in  Review  Problems:  To  make  W^^  the  selling  price 
must  be  |  of  cost,  or  \  of  24c.  =  30c.  But  30c.  is  WS  less  than  asking  price.  Since 
1615^  =  ^,  «  —  J  (=  5)  of  asking  price,  must  equal  selling  price,  and  \  of  asking  price 


SUGGESTIONS. 


equals  J  of  selling  price ;  therefore,  |  of  asking  price  equals  |  of  selling  price.    }  of 
30c.  =  36c.,  asking  price. 

353.  One  method  of  computation  well  taught  is  worth  more  than  all  the  others 
poorly  taught ;  and  all  special  methods  should  be  omitted  with  beginners. 

396.     Ex.  5  is  solved   by  formula    (g^  x  100^=  Rj.     Thus:   B  =  900, 

2-4  5 

P  =  $19.20,  t  ^  ih  =  i^  y-    Then  W|»  x  -\5^  x  ii&ift^  =  i  2% 

8. 
403-412.    The  five  cases  correspond  to  the  five  cases  of  simple  interest. 

426.  Both  Rules  and  Formulas  are  omitted  in  computations  of  Stocks  and 
Bonds,  because  they  are  the  simplest  applications  of  percentage. 

Prob.  17.  Another  Solution  :  Dividing  the  given  by  the  required  rate  %  gives 
the  rate.    .75  x  500  =  375 ;  or,  $375  is  the  price. 

427.  Letters  of  Credit,  that  is,  written  orders  on  which  partial  payments  are 
made  at  sight,  are  issued  to  travellers  in  all  or  nearly  all  civilized  countries  ;  thus 
affording  great  security  in  the  transportation  of  necessary  funds  for  travelling  ex- 
penses. The  value  of  such  letters,  of  course,  is  reckoned  according  to  the  principles 
governing  the  calculation  of  exchange. 

436.  The  table  on  p.  366  will  suggest  to  the  ingenious  teacher  a  great  variety 
of  interesting  problems. 

461.  Pupils  should  also  be  exercised  in  solving  Problems  in  Partnership  and 
Bankruptcy  by  Distributive  Proportion,  pp.  285-288. 

484.  Prob.  1.  Had  Simpson  paid  the  stipulated  sums  at  the  times  agreed 
tipon,  the  party  from  whom  he  bought  would  have  had  at  the  end  of  15  mo.  the 
interest  of  $500  for  15  mo.  ;  of  $600  for  9  mo. ;  and  of  $700  for  3  mo. ;  or,  at  6^,  he 
would  have  had  $37.50  +  $27  +  $10.50  =  $75  interest.  Simpson  then  should  pay  him 
$2700  ($500  +  $600  +  $700  +  $900)  at  such  a  time  be/ore  the  end  of  15  mo.,  that  at  the 
end  of  the  15  mo.  its  interest  at  6$?  would  equal  $75.  In  12  mo.  $2700,  at  6^,  will 
gain  $162 ;  and  will  gain  $75  in  ts\  of  12  mo.  =  5|  mo.  Simpson,  therefore,  should 
pay  $2700  in  15  mo.  —  5|  mo.  =  9 J  mo.^Ans. 

485.  Let  the  pupil  work  a  number  of  examples  by  selecting  both  the  first  and 
the  last  dates  as  focal  dates. 

609.  In  the  applications  of  Square  and  Cube  Roots,  no  demonstrations  are 
attempted. 

531.  Prob.  1 :  As  the  first  payment  is  made  at  the  beginning  ot  the  first  year, 
and  the  tenth  payment  at  the  beginning  oi  the  tenth  year,  the  entire  10  payments  are 
made  within  9  yrs.  and  1  da.  We  must,  therefore,  regard  the  annuity  as  having  10 
yrs.  to  run. 

567.  It  is  a  matter  of  regret  that  the  Metric  System  has  not  come  into  general 
use  in  the  United  States. 


^^^^5^ 


iniiis  sf  MrilJimBlic 


a 

H 


o 

H 

H 

H 

Oh 
g 
O 


3.   ITJVJT. 


CLASSIFICATION.  J 


OPERATION. 


TERMS. 


SIGXS. 


CLASSIFICATION.  ^ 


As  to  Object 

As  to  WJioleness 
of  their  Unit. 

As  to  their  Na- 
ture. 
As  to  the  Number 
ofJeindsofUn 


Us.  j 


4.  Concrete. 

5.  Abstract. 

6.  Integral. 

7.  Fractional. 

8.  Mixed. 
Like. 
Unlike. 

11.  Simple. 

12.  Compound 


■    9. 


Numeration  and  Notation. 

Addition. 

Subtraction. 

Slultiplica  tion. 

Division. 

Reduction. 

13.  Solution. 

14.  Problem. 

15.  Explanation, 

16.  Princiitle, 

17.  Example. 

18.  Analysis. 

19.  yjuie. 

Denominate  Nttmbers. 

Percentage. 

Ratio  and  Proportion. 

Involution  and  Evolution. 

Progressions. 

Mensuration. 


CHAPTER    I. 


SECTION    I.  — DEFINITIONS. 

1.  Arithmetic  is  the  science  of  numbers  and  the  art 
of  computation. 

3.  A  Wutuber  is  a  unit,  or  a  collection  of  units,  and 
answers  the  question  "How  many  ?  " 

3.  A  Unit  is  one,  or  a  single  thing;  as,  one,  one  dollar, 
OJie  house,  one  bushel,  one  peck,  one  half.  (5,  jstote,  and  150.) 

Numbers  are  classified  as  to  dfyject  into  Concrete  and  Abstract. 

4.  A  Concrete  Number  is  one  that  is  applied  to 
some  particular  object;  as,  three  books,  four  dollars,  five 
miles. 

5.  An  Abstract  Number  is  one  that  is  not  applied 
to  any  object ;  as,  three,  four,  five. 

Numbers  are  classified  as  to  their  umi  into  Integral,  Fractional,  and 

Mixed. 

6.  An  Integer  is  a  whole  number ;  as,  one,  five,  ten. 

7.  A  Fraction  is  a  number  expressing  one  or  more  of 
the  equal  parts  of  an  integer ;  as,  one  half,  five  tenths. 

8.  A  Mixed  Number  is  composed  of  an  integer  and 
a  fraction;  as,  six  and  one  half,  nine  and  five  tenths. 

Numbers  are  classified  as  to  their  nature,  into  Like  and  Unlike. 
IS 


INTEGERS.  13 

9.  lATze  Numbers  are  those  which  express  the  same 
kind  of  units  ;  as,  two  cents  and  four  cents,  six  hats  and 
one  hat 

10.  Unlike  Numbers  are  those  which  express  differ- 
ent kinds  of  units;  as,  two  cents  and  six  hats,  four  cents 
and  one  hat. 

Numbers  are  classified  as  to  the  number  of  kinds  of  units,  into 
Simple  and  Compound. 

11.  A  Simple  Number  is  a  number  expressing  one 
kind  of  unit ;  as,  four  pounds. 

13.  A  Compound  Number  is  a  number  expressing 
more  than  one  kind  of  unit;  as,  four  pounds  five  ounces. 

The  terms  used  in  computations  in  Arithmetic  are  Solution,  Problem, 
Explanation,  Principle,  Example,  Analysis,  and  Rule. 

13.  A  Solution  is  a  process  of  computation  used  to 
obtain  a  required  result. 

14.  A  Problem  is  a  question  for  solution. 

15.  An  Explanation  is  a  statement  of  the  reasons 
for  the  manner  of  solving  a  problem. 

16.  A  Principle  is  a  general  truth  upon  wbich  a 
process  of  computation  is  founded. 

17.  An  Example  is  a  problem  used  to  illustrate  a 
principle,  or  to  explain  a  method  of  computation. 

18.  An  Analysis  is  a  statement  of  the  successive 
steps  in  a  solution. 

19.  A  Mule  is  a  direction  for  performing  any  compu- 
tation. 


OUTLINE   OF   NUMERATION   AND   NOTATION, 


^    23.  ORIQiy. 

24.  CHARACTERS. 

86.  Significant. 
25.  Names.   ■{ 

27.  Insignificant. 

i   29.  Simple. 
28.  Values.    ■{ 

\    30.  Xoco/. 

33,  Units  or  Ones. 

^ 

31.  J'/oce 

33.  TeTW. 

or 

. 

34.  Hundreds. 

< 

TERMS. 

&c.,  <Sx.,  &c. 

■  36.  fT^iite. 

01 

37.  Thousands. 

35.  Period. 

38.  i/i/Zto^w. 

39.  BUlions. 

. 

tfec,  cfec.,  tfec 

4:0.  TABLE. 

41.  PRINCIPLES. 

^    42.  RULES. 

r    44.  ORIGIN. 

46.  i^amcs. 

^ 

'    47.  ^to7i«. 

45.  CHARACTERS.    . 

48.  Repeated. 

Faluetf.    ■ 

49.  FoUowing. 

50.  Preceding. 

CO 

51.  Between. 

^   52.  Horizontal  Lint. 

.63.  17S^«. 

SECTION    II. 


i! 


(sr^i 


g-^ 


?MBMER!mit®l«  JIBi©  NQ^T^llOM  -f 


(S-^-^ 


'6' 


<^v2) 


If 


20.  Numeration  is  a  method  of  reading  numbers 
represented  by  characters. 

21.  Notation  is  a  method  of  writing,  or  representing 

numbers  by  characters. 

There  are  two  methods  of  reading  and  writing  numbers,  viz.  :  the 
Arabic  and  the  Roman. 

22.  In  the  Arabic  Method  numbers  are  expressed 
by  means  of  ten  characters. 

23.  The  Arabic  method  of  expressing  numbers  had  its 
origin  in  India,  but  was  introduced  into  Europe  by  the 
Arabs. 

24.  The  Characters  used  in  this  system  of  expressing 
numbers  are  cdlle^  figures. 

25.  The  figures  used  and  their  names  are  as  follows: 
Printed,       12345678       90 

Written.        /2SJ^J^7§pO 
Named,        One,  Two,  Three,  Four,  Five,  Six,  Seven,  Eight,  Nine,  Naught. 

26.  The  first  nine  are  called  significant  figures, 
because  each  always  expresses  a  number.  They  are  some- 
times called  digits,  from  the  Latin  digitus,  a  finger;  it 
being  once  customary  to  count  upon  the  fingers. 


16  INTEGERS. 

27.  The  figure  0,  naught,  is  also  called  cipher  and  zero, 
and  signifies  7io  value  or  nothing, — hence  insiffnificant, 

28.  Ihe  Value  of  a  figure  is  its  power  to  express 
quantity. 

Figures  have  two  kinds  of  value,  viz. :  Simple  and  LoccU. 

29.  The  Simjyle  value  of  a  figure  is  the  quantity  which 
it  expresses  when  it  is  alone.  Thus  5  equals  five,  8  equals 
eight. 

30.  The  Local  value  of  a  figure  is  the  quantity  which 
it  expresses  by  occupying  a  place  in  a  number.  Thus  in  50, 
S  ec^usds  fifty  ;  in  800,  S  equals  eight  hundred. 

In  the  units'  place  of  a  number  the  simple  and  local  values  of  a 
figure  are  the  same. 

The  terms  used  in  expressing  numbers  by  figures  are  places  or 
orders,  and  periods. 

31.  The  JPlace^  or  Ofder^  of  a  figure  is  its  position  in 
a  number.  Thus  in  365,  5  is  in  the  1st  Place  ;  0,  in  the 
2d ;  and  3,  in  the  3d. 

32.  The  name  of  the  first  place,  or  order  is  Units,  or 
Ones. 

Tlie  greatest  number  of  units  expressed  hy  one  figure  is  9. 

33.  The  name  of  the  second  place  or  order  is  Tens. 

When  two  figures  are  used  to  express  a  number,  the  left-hand  one 
expresses  tens  and  the  right-hand  one  ones.     Thus, 


10  means 

11 

13 

13 

14 

15 


ten  0  ones ten, 

"   lone   ("endelfen,"         — one     and  ten)  eleven, 
"■   2  ones  ("  twelif,"  — two    and  ten)  twelve. 

"3  "  ("  thir  and  teen," — three  and  ten)  thirteen. 
'*  4  "  ("  four  and  teen," — four  and  ten)  fourteen. 
"5    "     ("  fif  and  teen,"    — five     and  ten)  fifteen. 


16,  sixteen ;  17,  seventeen  ;  18,  eighteen  ;  19,  nineteen. 


KUMERATION^     AND     N^  0  T  A  T  I  0  N^ 


17 


20  means  2  tens  0  ones  ("  tween  ty,"  two  tens)  twenty. 

21  "      2    "     lone  ("  tween  ty"  and  one, — two  tens  and  one) 

twenty-one. 


22,  twenty-two. 

23,  twenty-three. 


24,  twenty-four. 

25,  twenty-five. 


26,  twenty-six. 

27,  twenty-seven. 


28,  twenty-eight. 

29,  twenty -nine. 


The  greatest  number  tUat  can  he  expressed  Jyy  two  figures  is  99. 
34.  The  name  of  the  third  place  or  order  is  Mundreds, 

When  three  figures  are  used  to  express  a  number,  the  left-hand 
one  expresses  hundreds,  the  next,  tens,  and  the  next,  ones. 


100  means  1  hundred  0  tens  0  ones 

101  "1        "         0    "     1  one 
110      "      1        "  Iten   Oones 
200      "      2  hundreds  0  tens  0    " 


one  hundred, 
one  hundred  one. 
one  hundred  ten. 
two  hundred. 


The  greatest  number  that  can  he  represented  hy  three  figures  is  999. 


Write  in  figures 

1.  One  hundred  twenty-five. 

2.  Four  hundred  fifty-nine. 

3.  Five  hundred  eighty, 

4.  Six  hundred  eleven. 

5.  Eight  hundred  seventy. 

6.  Two  hundred  ninety-nine. 

7.  Seven  hundred  forty-four. 

8.  Three  hundred  thirty-five. 

9.  Nine  hundred  nineteen. 
10.  Five  hundred  five. 


the  following: 

11.  Seven  hundred. 

12.  Three  hundred  thirty. 

13.  Eight  hundred  one. 

14.  Six  hundred  twelve. 

15.  Fifty-eight. 

16.  Seventy-nine. 

17.  Nine  hundred. 

18.  One  hundred  ninety. 

19.  Seven  hundred  ten. 

20.  Nine  hundred  ninety. 


35.  A  Pei*iod  consists  of  three  places  or  orders. 

36,  In  expressing  numbers,  {hQ  first  three  orders  or  places 
are  regarded  as  forming  a  group,  called  the  Period  of 
Units. 

2 


18  IKTEGEKS. 

TABLE    OF    UNITS. 

Dnits. 

10  units  (1)  make  1  ten       10 

10  te,n8  "     1  hundred 100 

10  hundreds     "     1  thousand 1  000 

37.  The  second  group  of  three  orders  or  places  from  the 
right  is  called  the  JPeri^d  of  Thousands. 

I 

TABLE    OF    THOUSANDS.  J     | 

10  thousands  make  1  ten-thousand 10  000 

10  ten-thousands             **     1  hundred-thousand  .    .        .        100  000 
10  hundred-thousands    *'     1  million 1  000  000 

123456  is  composed  of  123  thousands  456  units,  or  of  two  periods,  and 
is  read  123  thousand  456. 

300004  is  composed  of  300  thousands  4  units,  or  of  two  periods,  and 
is  read  300  thousand  4. 

4125  is  composed  of  4  thxmsands  125  units,  or  of  two  periods,  and 
is  read  4  thousand  125. 

38.  The  third  group  of  three  orders  or  places  from  the 
right  is  called  the  Period  of  Millions. 

TABLE    OF    MILLIONS.  I      i     | 

s     ^     J 

10  millions  make  1  ten-million 10  000  000 

10  tenrmillions            "     1  hundred-million      .     .     .        100  000  000 
10  hundred-mUlions    "     1  billion 1  000  000  000 

451219149  is  composed  of  451  millions  219  thousands  149  units,  or  of 
three  periods,  and  is  read  451  million  219  thousand  149. 

75000100  is  composed  of  75  millions  100  units,  or  of  three  periods, 
and  is  read  75  million  100. 

3040000  is  composed  of  3  millions  40  thousand,  or  of  three  periods, 
and  is  read  3  million  40  thousand. 


KUME  RATION     AND     NOTATION, 


19 


39.  The  fourth  group  of  three  orders  or  places  from  the 
right  is  called  the  JPeriod  of  Billions. 

TABLE    OF    BILLIONS.      |     i     |     .| 

10  biUions  make  1  ten-billion To  000  000  000 

10  ten-billions            "      1  hundred-billion    ...        100  GOO  000  000 
10  hundredMllions   "      1  trillion 1  000  000  000  000 

JSliSoOOOOO  is  composed  of  413  billions  123  millions,  or  of  fo7zr 
periods,  and  is  read  412  billion  123  million. 

'^OOOoioOOO  is  composed  of  90  billioTis  50  thomands,  or  of  four 
periods,  and  is  read  90  billion  50  thousand. 

40.  The  names  of  the  Orders  of  Units  and  the  names 
of  the  Periods  are  given  in  the  following 

TABLE. 


Periods. 


SIXTH         rrPTH       rOTTRTH       THIBD       SECOND        ITRST 
PERIOD.     PERIOD.     PERIOD.      PERIOD.     PERIOD.     PERIOD. 


Nanjes  of  Periods.  ■ 


Orders  of  Units. 
Number. 


<§ 


of 


of 


of 


of 


of 


ttj    !>l    Q> 


III 


t^  E><  5 


^1^ 


400    040    004    444    440 


of 


404 


This  number  is  read  400  quadrillion  40  trillion  4  lUlion  4M  million 
440  thousand  404. 

The  Periods  above  Quadrillions  are  Quintillions,  SeTtillions,  Septil- 
lions,  Octillions,  Nonillions,  DecilUons,  Undedllions,  Duodecillions, 
Tredecillions,  Quatuordecillions,  Quindecillions,  Sexdecillions,  SeptendeciJf- 
lions,  OctodeciUions,  NovendeciUioTis,  Vigintillions,  &c. 


20 


IITTEGERS. 


PRINCIPLES    OF    NUMERATION    AND    NOTATION. 

41 .  First.  Ten  units  of  any  order  in  a  number  are  always  equal 
to  one  unit  of  the  next  higher  order. 

Second.  Removing  a  figure  one  place  to  the  left,  increases  its  value 
tenfold.  Removing  a  figure  one  place  to  the  right,  diminishes  its  value 
tenfold. 

Third.  The  name  and  value  of  the  units  represented  hy  a  figure  in 
a  number  are  always  those  of  its  order  in  that  number. 

Fourth.  The  absence  of  units  in  any  order  in  a  number  is  denoted  "by 


RULES    FOR    NUMERATION    AND    NOTATION. 

42.  Numeration". — Beginning  at  the  right,  separate  the 
numbers  i?ito  periods  of  three  figures  each.  TJien  read  the 
number  in  the  first  period  at  the  left,  adding  the  na^ne  of  the 
period  J  do  the  same  with  each  period  in  order  toivard  the 
right,  omitting  to  name  the  units  period  and  any  period  or 
periods  composed  only  of  ciphers. 

Notation. — Beginning  at  the  left,  write  the  hundreds, 
tens,  and  ones  of  each  period  in  their  proper  order,  filling  all 
vacant  orders  and  periods  with  ciphers. 


EXER  CISES. 

Eead  the  following  numbers : 

1.  125  623 

6.  423  152  391 

11.    2  648 

2.  400  119 

7.   72  110  110 

12.    9  999 

3.   50  090 

8.   95  486  000 

13.   14  150 

4.   62  124 

9.  172  000  400 

14.   40  000 

5.  142  205 

10.  909  412  613 

15.  425  003 

16.  144  044  014  005  009 

21.  9  00 

9  009  009  009  009 

17.  75  000  000  000  075 

22.  41  00 

0  041  000  041  000 

18.     48  370  490  563 

23.  3  00 

0  000  000  000  000 

19.  49  123  436  000  000 

24.    6 

2  410  000  443  000 

20.  11  010  001 

936  127 

25.    41 

3  125  679  456  199 

NUMEEATIOIf     AND     NOTATIOK.  21 

Write  the  following  numbers  in  figures : 

1.  Twenty-two  thousand  seven  hundred  sixty-five. 

2.  Eighty  thousand  two  hundred  one. 

3.  Thirty  thousand  thirty. 

4.  Four  hundred  ten  thousand  two  hundred  five. 

5.  Ninety  thousand  one. 

6.  Eight  hundred  thousand  six  hundred  sixty-nine. 

7.  Nine  hundred  thousand  one. 

8.  Five  hundred  thousand  fifty. 

9.  One  hundred  million  ten  thousand  one. 

10.  Ninety-one  million  seven  thousand  sixty. 

11.  Seventy  million  four. 

12.  Seven  hundred  million  ten  thousand  one. 

13.  One  billion  one  million  forty. 

14.  Forty  billion  two  hundred  thousand  five. 

15.  Seven  hundred  twenty-six  billion  fifty  million  one 
thousand  two  hundred  forty-three. 

ROMAN     NOTATIOIV. 

43.  In  the  Roman  3Iethod  numbers  are  expressed 
by  means  of  seven  characters. 

44.  The  Roman  method  of  expressing  numbers  had  its 
origin  in  Rome. 

45.  The  Characters  used  in  this  system  of  expressing 
numbers  are  seven  Letters, 

46.  The  letters  used  are  the  following : . 

I,         V,         X,  L,  C,  D,  M. 

One,  Five,  Ten,  Fifty,      One  hundred,  Five  hundred,  One  thousand. 


»55  INTEGERS. 

47.  When  used  alone  each  letter  has  a  fixed  value. 

48.  When  a  letter  is  repeated,  its  value  is  repeated. 
Thus,  III  are  3,  XX  are  20,  COO  are  300,  MM  are  2000. 

49.  When  a  letter  folloivs  one  of  greater  value,  their 
sum  is  the  number  expressed.     Thus,  VI  are  6,  LX  are  60. 

50.  When  a  IqUqt  precedes  one  of  greater  value,  their 
difference  is  the  number  expressed.  Thus,  IV  are  4, 
XL  are  40. 

51.  When  a  letter  is  placed  between  two  others  ol 
greater  value,  the  difference  hetioeen  it  and  their  sum  is  the 
number  expressed.     Thus,  XIX  are  19,  XIV  are  14. 

53.  When  a  horizontal  line  is  placed  over  a  letter, 
its  value  is  increased  a  thousand  times.    Thus  V  is  5000. 

53.  Uses, — Roman  notation  is  not  convenient  for  arith- 
metical computations,  but  is  used  principally  in  marking 
dials,  numbering  chapters,  sections,  &c.,  in  books,  and  in 
Writing  physicians'  prescriptions. 


EX  ER  CISE8. 

Write  the  following 

numbers  in  letters : 

1.     7 

9.  88 

17.  109 

25.  482 

33.  2300 

2.  14 

10.  94 

18.  245 

26.  375 

34.  5050 

3.  28 

11.  73 

19.  537 

27.  160 

35.  4875 

4.  56 

12.  59 

20.  199 

28.  543 

36.  1876 

5.  50 

13.  60 

21.  125 

29.  215 

37.  1250 

6.  75 

14.  85 

22.  150 

30.  312 

38.  2275 

7.  96 

15.  63 

23.  260 

31.  219 

39.  2116 

8.  99 

16.  89 

24.  384 

32.  420 

40.  5000 

OUTLINE    OF    ADDITION. 


TERMS. 


55.  Addends,  or  Parts, 

56.  Sum,  or  Amount, 


o 


57.  SIGNS. 


61.  TABLE. 


58.  Of  Addition, 

59.  Of  Equality. 

60.  Of  Dollars. 


Q 

Q 


PRINCIPLES. 


CASES. 


62.  Like  Numbers. 

63.  Like  Orders, 


64.  L 


65.  IL 


When  the  sum  of 
the  units  of  each 
order  is  less 
than  10. 

Wh£n  the  svm  of 
the  units  of  any 
order  equals  or 
exceeds  10. 


66.  RULE. 


SECTION   III 

<G)©(?)- 


ilLl>)I)jI^JlQjM 


.^j,^ 

54.  Addition  is  the  process  of  uniting  two  or  more 

numbers  to  form  one  number. 

The  Terms  used  in  Addition  are  Addends,  or  Parts,  and  Sum,  or 
Amount. 

55.  The  Addends,  or  JParts,  are  the  numbers  to  be 
added. 

56.  The  Sum,  or  Amount,  is  the  result  obtained  by 
Addition. 

57.  Signs  are  characters  used  for  abbreviating  expres- 
sions. 

58.  The  Sign  of  Addition  is  an  erect  cross  called 
plus  ( + ),  and  when  placed  between  numbers  signifies  that 
they  are  to  be  added.  Thus,  3  +  2  means  that  3  and  2  are 
to  be  added,  and  is  read  "  3  plus  2." 

The  term  plus,  Latin,  signifies  more,  or  added  to.  « 

59.  The  Sign  of  Equality  is  two  short,  horizontal, 
parallel  lines  (=),  and  when  placed  between  numbers  sig- 
nifies that  they  are  equal.  Thus,  3+2  =  5  means  that 
3  and  2  are  equal  to  5,  and  is  read  "  3  plus  2  equals  5." 

60.  The  Sign  of  Dollars  is  a  capital  S  with  two 
vertical  marks  across  it  (I),  and  when  placed  before  a  num- 
ber expresses  dollars.    Thus,  $120  means  120  dollars. 


ADDITION 


25 


61.    ADDITION    TABLE. 


0    12 
111 

12    3 

ONE. 

3  4   5    6    7    8    9 
1111111 

4  5    6    7    8    910 

SIX. 

01234567    8    9 
6666666666 

6    7    8    910111213  1415 

0    12 
2    2    2 

TWO. 

3    4    5    6    7    8    9 
2    2    2    2    2    2    2 

SEVEN. 

0123456789 

7777777777 

2    3    4 

5    6    7    8    91011 

7    8    910111213  1415  16 

THREE. 

0123456789 
3333333333 

EIGHT. 

0123456789 

8888888888 

3    4   5 

6    7    8    9101112 

8    9  10111213  1415  16  17 

0    12 
4   4   4 

FOUR. 

3  4    5    6    7    8    9 

4  4   4   4   4   4   4 

NINE. 

0123456789 
9999999999 

4   5    6 

7    8    910  111213 

910111213  1415  16  1718 

0    12 
5    5    5 

FIVE. 

3    4    5    6    7    8    9 
5    5    5    5    5    5    5 

TEN. 

0123456789 
10  10  10  10  10  10  10  10  10  10 

5    6    7 

8    9  10111213  14 

101112131415  16  171819 

26  INTEGEBS 


;f  ©i^at^xTGi^ci^eX 


Example. — James  has  two  hats  and  John  has  3.  How 
many  have  both  ? 

Solution. — Since  2  hats  and  3  Tiats  are  6  hats,  they  both  have  5  hats. 
Or,  2  hats  and  3  hats  a/re  6  hats.  TJierefore.  they  loth  have  5  hats. 
Or,  They  both  have  6  hais,  because  2  fiats  and  3  hois  are  5  hats. 

1.  Willie  found  4  marbles  and  bought  7.  How  many  had 
he  then  ?     7  and  4  are  how  many  ?     9  and  2  ? 

2.  Samuel  has  5  walnuts  and  James  gives  him  4  more. 
How  many  has  he  then  ?  How  many  are  5  and  4  ?  4  and  5  ? 
2  and  2  and  5?  3  and  2  and  4?  1+4+4=?  1  +  3  +  5=? 
2  +  2  +  2  +  3  =  ? 

3.  Charles  bought  7  almonds  at  one  store  and  10  at 
another.  How  many  did  he  buy  in  all  ?  7  +  10  =  ? 
10  +  7  =  ?  7  +  5  +  5  =  ?  2  +  8  +  7  =  ?  6  +  7+4  =  ? 
3+7  +  7  =  ?     9  +  3  +  5  =  ? 

4.  If  a  pound  of  sugar  costs  9  cents  and  an  orange  6  cents, 
what  will  both  cost  ?  6  cents  and  9  cents  are  how  many 
cents  ? 

5.  Jennie's  father  paid  8  dollars  for  her  muff,  7  dollars 
for  her  tippet.     How  much  did  he  pay  for  both  ?    7  +  8  =  ? 

6.  If  a  house  has  7  windows  in  the  front  and  6  in  the 
rear,  how  many  windows  has  it?  6  +  7  =  ?  4  +  2  +  5 
+  2  =  ? 

7.  Mary  paid  8  cents  for  an  inkstand  and  5  cents  for  a 
pencil.  How  much  did  she  pay  for  both  ?  5  cents  + 
8  cents  =  ? 


ADDITION. 


27 


How  many  are 


8.10menand7men?,13.5  +  2  +  3? 
9.  8  boys  and  9  boys  ?  14.5  +  3  +  1? 

10.  6  girls  and  8  girls  ? 

11.  5  tops  and  9  tops  ? 

12.  3  pins  and  8  pins  ? 


15.3  +  3  +  3? 

16.4  +  4  +  4? 

17.5  +  5  +  5? 


18.  7  toys  and  7  toys? 

1 9.  6  miles  and  8  miles  ? 

20.  8  cups  and  4  cups  ? 

21.  7  cents  and  5  cents  ? 

22.  6  figs  and  5  figs  ? 


23.  How  many  are  8  chairs  and  4  dollars  ? 

We  cannot  add  8  chairs  and  4  dollars.     For  8  chairs  +  4  dollars  = 
neither  12  chairs  nor  13  dollars.     Hence, 

63.   Only  abstract  numbers,  or  like  concrete  numbers,  can 
be  added. 

24.  How  many  are  3  tens  and  9  thousands  ? 

We  cannot  add  3  tens  and  9  thousands.     For  3  tens  +  9  thousands 
=  neither  12  tens,  nor  12  thousands.     Hence, 

63.  Only  like  orders  of  units  can  be  added. 


fWfriiien  ^^erci^e^-f 


CASE    I. 

64.  When  the  sum  of  all  the  units  of  each 
order  is  less  thaii  10> 

Example.— What  is  the  sum  of  11,  431,  2245,  3112  ? 

SOLUTION. 

11   = 1  ten    1  one 

431   = 4  hundreds  3  tens  1  one 

2245   =  2  thousands  2  hundreds  4  tens  5  ones 
3112  =  3  thousands  1  hundred    1  ten    2  ones 

Amount.      5799  =  5  thousands  7  hundreds  9  tens  9  ones 


Addends. . 


28 


IlS^TEGEKS. 


Explanation. — Since  only  like  orders  can  be  added,  it  is  most  con- 
venient to  write  like  orders  in  the  same  column. 

We  first  add  the  ones,  1,  1,  5,  and  3,  and  find  tlieir  sum  to  be  9, 
which  we  write  in  the  ones'  order  of  the  amount. 

We  then  add  the  tens,  1,  3,  4,  and  1,  and  find  their  sum  to  be  9, 
which  we  write  in  the  tens'  order  of  the  amount. 

We  then  add  the  hundreds,  4,  3,  and  1,  and  find  their  sum  to  be  7, 
which  we  write  in  the  hundreds'  order  of  the  amount. 

We  then  add  the  thousands,  3  and  3,  and  find  their  sum  to  be  5, 
which  we  write  in  the  thousands'  order  of  the  amount. 

The  result,  5799,  is  the  sum  required. 


rJtOBLEMS. 


123 

12 

132 

.-67 


(2.) 

410 

30 

240 

680 


(3.) 
432 
201 

222 
855 


(4.) 

210 
135 

42 

387 


(5.) 
1234 
1211 
4524 

6969 


(6.) 

231 

2213 

6404 


(7.) 
210414 
337052 
142513 


8848    689979 


8.  Add  6404,  2213,  and  1231.  A?is.   9848. 

9.  Add  3472,  4513,  1001,  and  1000.      Ans.   9986. 

10.  Add  10,  12,  21,  32,  11,  and  3.  Ans.   89. 

11.  Add  1321,  2312,  1241,  and  3102.  Ans.  7976. 

12.  Add  3131,  2112,  3421,  and  1003.  Ans.  9667. 

13.  Add  10102,  4030,  634,  11,  and  2.  Ans.  14779. 

14.  John  gave  10  cents  for  a  spelling-book,  12  cents  for  a 
writing-book,  15  cents  for  a  slate,  20  cents  for  pens  and  ink, 
and  31  cents  for  an  Arithmetic.  How  much  did  he  give 
for  all?  Ans.  88  cents. 

15.  One  horse  cost  120  dollars,  another  115  dollars,  one 
harness  31  dollars,  the  other  30  dollars,  and  the  carriage 
200  dollars.     How  much  did  all  cost?     Ans.  496  dollars. 

16.  Gave  3216  dollars  for  land,  2150  for  a  house,  1502  for 
stock,  and  2121  for  a  barn.     How  much  did  all  cost  ? 

Ans.  8989  dollars. 


ADDITIOi?^. 


f  ©ral^xfGl^ciXGX 


Example. — How  many  are  16  and  9  ?     (Written.) 

SOLUTION.  EXPLANATION. 

)        9      r6  ones  and  9  ones  are  15  ones  =  1  ten 
1st  step.  >■     16     •<      and  5  ones. 

)        5      (  Write  the  5  ones  in  ones'  order  of  sum. 

)        9      (  Add  the  1  ten  to  tens'  order  of  parts. 
2d  step.  >•      16      •<  1  ten  and  1  ten  are  2  tens. 

)      25      (  Write  the  2  tens  in  tens'  order  of  sum. 

Ans.  25. 
Example. — How  many  are  16  and  9  ?    (Oral.) 
Explanation.— Add  the  right-hand  figures,  9  and  6.    Their  sum  is 
15 ;  retain  the  5  in  the  mind,  and 

Add  the  1  to  the  left-hand  figure,  1.     Their  sum  is  3. 
Result,  25. 

PJBOBIjEMS. 

1.  How  many  are   18  and   9?     15  and  6?    17  and  5? 
13  and  8?    16  and  4?     19  and  7? 

2.  How  many  are   24  and   7?     28  and  5  ?     26  and  9? 
25  and  8  ?     29  and  5  ?     27  and  8  ? 

3.  How  many  are  36  and  6  ?     32  and  9  ?    38  and  5  ? 
35  and  7?    37  and  4?     39  and  8? 

4.  How  many  are  47  and  8  ?    44  and  6  ?    49  and  7  ? 
42  and  9  ?    46  and  5  ?    48  and  9  ? 

5.  How  many  are  57  and  4  ?    59  and  7  ?     55  and  6  ? 
53  and  8?    57  and  9?     58  and  5? 

6.  How  many  are  62   and   9?     65  and  8  ?    67  and  7? 
64  and  6?     68  and  9?    69  and  9? 


30 


INTEGERS. 


7.  How  many  are  75  and  7  ?  73  and  9  ?  77  and  5  ? 
78  and  4  ?     79  and  6  ?  76  and  9  ? 

8.  How  many  are  83  and  9?  87  and  6?  85  and  8? 
86  and  7  ?     88  and  7  ?  89  and  9  ? 

9.  How  many  are  91  and  9?  95  and  8?  98  and  6? 
94  and  7  ?     99  and  5  ?  93  and  9  ? 

Example. — How  many  are  43  and  39  ? 

Explanation. — See  tLe  sum  of  right-hand  figures  (12),  retain  in 
the  mind  the  ones  of  this  sum  (2). 

See  the  sum  of  left-hand  figures  (7),  increase  this  by  1  (8),  because 
the  sum  of  the  right-hand  figures  exceeded  9.    Ans.  82. 

10.44  +  37  =  ?  45  +  36  =  ?  42  +  39  =  ?  48  +  34  =  ? 
11.51  +  19  =  ?  56  +  18  =  ?  57  +  15=?  59  +  13  =  ? 
12.62  +  18  =  ?  65  +  13  =  ?  64  +  25  =  ?  69  +  15  =  ? 
13.71  +  18  =  ?  75  +  25  =  ?  74  +  16  =  ?  79  +  11  =  ? 

14.  12  +  8  +  9  +  7  +  8  =  ?  15  +  9  +  8  +  9  +  15  =  ? 

15.  14  +  7  +  9  +  12  +  18  =  ?  17  +  3  +  40  +  20  +  9  =  ? 

16.  13  +  9  +  11  +  44  +  33  =  ?  16  +  10  +  4  +  9  +  12  =  ? 
17.24  +  7  +  9  +  60  +  50  =  ?  25  +  15  +  25  +  15  =  ? 
18.  1  +  2  +  3  +  4  +  5  +  6  +  7  +  8  +  9  +  10  +  11  +  12  =  ? 


D. 


tWritten  ^Xer  eigesT-t 


CASE    II. 

65,  When  the  sum  of  all  the  units  of  each 
order  equals  or  exceeds  10, 

Example.— Find  the  sum  of  314,  591,  856,  721,  and  672. 


ADDITION.  31 

SOLUTION.  Explanation. — We  write  tlie  numbers  so  that  units 

q-i  ^  of  the  same  order  stand  in  tlie  same  column,  and  thus 

form  as  many  columns  of  figures  as  there  are  different 


591 

orders  of  units. 

856  Then  we  add  first  the  right-hand  or  units'  column, 

721  beginning  at  the  top  and  adding  downward. 

672  4:  +  l  +  6  +  l+2  =  14:,  OT  1  ten  and  4  ones. 

Write  the  4  ones  directly  under  the  ones,  and  add 

^^^^  the  1  to  the  tens. 

In  like  manner  we  add  the  tens'  column,  1  +  9  +  5  + 

2  +  7  and  the  1  ten  which  we  obtained  by  adding  the  ones'  column, 

making  in  all  35,  or  3  hundreds  and  5  tens. 

We  write  the  5  tens  directly  under  the  column  of  tens  and  add  the 

2  hundreds  to  the  column  of  hundreds. 

The  hundreds'  column  3  +  5  +  8  +  7  +  6  and  the  3  hundreds  from  the 

tens  equal  31  hundreds,  or  3  thousands  and  1  hundred. 

Write  the  1  hundred  directly  under  the  column  of  hundreds  and 

the  3  thousands  in  the  thousands'  order  of  the  result. 

This  completes  the  work  and  gives  us  3154  for  the  sum. 

We  should  obtain  the  same  result  by  beginning  at  the  bottom  and 
adding  upwards.  It  is  mainly  by  repeated  additions,  in  this  manner, 
that  we  can  prove  the  correctness  of  the  result  obtained. 

66.  EuLE. — Arrange  the  numbers  to  he  added  so  that  the 
same  order  of  units  shall  stand  in  the  same  column. 

Find  the  sum  of  the  ones'*  column  and  place  the  right-hand 
figure  of  the  sum  directly  under  the  ones  and  add  the  num- 
ber expressed  hy  the  left-hand  figure  or  figures,  if  any^  to  the 
tens'  column. 

Add  in  the  same  manner  each  column  in  order,  always 
placing  directly  under  the  column  added  the  right-hand 
figure  of  its  sum,  and  adding  the  number  expressed  by  the 
left-hand  figure  or  figures,  if  any,  to  the  next  column,  talcing 
care  also  to  write  at  the  left  the  whole  of  the  last  sum  found. 


32 

IIJ^TEGERS. 

Pit  OBI.  JEMS 

'. 

(1.) 

(2.) 

(3.) 

(4.) 

(5.) 

(6.) 

(7.) 

142 

344 

796 

9 

98 

8 

4501 

177 

792 

438 

976 

472 

69 

2437 

802 

671 

29 

854 

9 

987 

684 

1121 

1807 

1263 

1839 

579 

1064 

7622 

8.  Add  873254,  289094,  627380,  and  456987. 

Ans.  2246715. 

9.  Add  35709,  50804,  90006,  and  47103.  Ans,  223622. 

10.  Add  263,  5827,  39001,  70126,  and  85.  Ans.   115302. 

11.  Add  92,  924,  9857,  95064,  and  572.  Ans.   106509. 

12.  Add  3597,  82,  649,  30548,  and  762.   Ans.   35638. 

13.  Add  4689,  8694,  75,  64982.        Ans.   78440. 

What  is  the  sum 

14.  Of  5  +  15  +  25  +  35  +  45  +  55  +  65?  Ans.   245. 

15.  Of  4+54+64  +  74  +  84  +  94  +  44?  Ans.   418. 

16.  Of  23  +  33  +  43  +  53  +  63  +  73  +  83?  Ans,   371. 

17.  Of  32  +  42  +  52  +  62  +  72  +  82  +  92?  Ans.   434. 


(18.) 

78243 

8145 

94067 

72 

21698 

202225 

313332 

717782 


(19.) 

43285 

16007 

990 

8622 

73084 

141988 

265888 

549864 


(20.) 

43 

437 

6908 
97360 

5033 
109781 
345678 

565240 


(21.) 

9 

89 

789 

6789 

56789 

•64465 

89658 

218588 


(22.) 
53102 

8967 

406 

12 

8 

62495 

95642 

220632 


ADDITION.  33 

23.  Add  six  hundred  forty- two,  three  thousand  one  hun- 
dred twenty-four,  seventy-nine  thousand  nine  hundred  six, 
eight  hundred  twenty-four,  seven  hundred  five,  and  forty- 
seven  thousand  twenty-eight.  Ans.  132229. 

24.  Find  the  sum  of  six  million  sixty  thousand  six,  seven 
million  nine  hundred  fifty  thousand  ninety-nine,  ten  million 
nine  thousand  eight  hundred  seven,  and  three  hundred 
sixty-seven  thousand  forty-five.  Ans,  24386957. 

25.  Add  seventy  thousand  four  hundred  fifty-three,  five 
million  eight  hundred  six  thousand  twenty-eight,  eighty 
million  ninety-seven  thousand  nine,  and  twenty-five  million 
seven  hundred  thousand.  Ans.  111673490. 

26.  A  man  owns  a  shop  worth  $4000,  a  house  and  lot 
worth  $7500,  bank  stock  worth  $1800,  insurance  stock 
worth  $800,  a  carriage  worth  $450,  and  two  horses  worth 
$350.     How  much  are  all  worth  ?  Ans.  $14900. 

27.  Mr.  B's  State  taxes  are  $91;  County  taxes,  $321; 
School  taxes,  $421 ;  Borough  taxes,  $375 ;  and  Poor  tax, 
$39.     What  is  the  whole  amount  of  taxes  ?      Ans.  $1247. 

28.  A  merchant  deposits  in  bank  on  Jan.  1,  $300; 
Jan.  3,  $1716;  Jan.  4,  $5217;  Jan.  9,  $16714;  Jan.  10, 
$516;  and  Jan.  11,  $7060.  How  much  in  all  has  he 
deposited?  Ans.  $31523. 

29.  A  drover  bought  350  sheep  from  one  farmer,  425 
from  a  second,  750  from  a  third,  400  from  a  fourth,  1616 
from  a  fifth.    How  many  did  ha  buy  from  aU  ? 

Ans.  3541. 

30.  From  the  savings  bank  0  drew  out  the  following 
sums:  $210,  $35,  $48,  $16,  $25,  $45,  $24,  $16,  $13,  $75,  $84, 
$7,  and  $16.    How  much  did  he  draw  out  in  all  ? 

Ans.  $614. 


34  INTEGERS. 

31.  A  merchant  "checked  out"  of  his  hank  $215,  $3125, 
$21620,  $17540,  $2063,  $216,  $257,  $301,  $51,  and  $1617. 
How  much  in  all  did  he  check  out  ?  Ans,  $47005. 

32.  A  man  after  "  checking  out "  $947,  found  that  he  had 
remaining  in  bank  $369.     How  much  had  he  at  first  ? 

A71S.  $1316. 

33.  After  "  checking  out"  $30,  $40,  $9,  and  $25,  a  grocer 
found  that  he  had  $711  remaining  in  bank.  How  much 
had  he  at  first  ?  Ans.  $824. 

34.  A  college  pays  its  president  $2500  per  year,  its  pro- 
fessors $16000,  its  assistant  professors  $4500,  and  its  tutors 
$4800.    How  much  does  it  pay  annually  for  instruction  ? 

Ans,  $27800. 

35.  My  income  from  rent  of  warehouse  is  $5750,  from  one 
dwelling-house  $1200,  from  another  $900,  from  bank  stock 
$3500,  from  insurance  stocks  $1500,  from  my  share  in  a 
steamboat  $1643,  from  interest  on  money  loaned  $673. 
What  is  my  entire  income  ?  Ans.  $15166. 

36.  G's  expenses  for  1  year  were :  rent  $1200 ;  taxes 
$225  ;  insurance  on  furniture  and  household  goods  $65 ; 
household  expenses  $3652;  James  and  Mary's  school  biUs 
$945 ;  clothing  for  self  and  family  $2250 ;  traveling  ex- 
penses and  hotel  bills  $742.    Find  the  sum.     Ans.  $9079. 

37.  A  invests  $2516  in  a  farm  ;  B  $16525  in  a  mill ; 
C  invests  $21621  in  a  steamboat ;  and  D  as  much  as  A,  B, 
and  C  together  in  a  cotton  factory.  How  much  do  they  all 
invest?  Ans,  $81324. 

38.  A  merchant  bought  of  a  jobber,  prints  amounting  to 
$3520  ;  muslins  $3962  ;  hnens  $1733  ;  silks  $2375 ;  woolen 
goods  $7675.     How  much  is  the  biU?  Ans.  $19265. 


ADDITIOIS^.  35 

39.  A  farmer  has  16  horses,  11  colts,  32  milch  cows, 
11  steers,  24  yearlings,  15  calves,  321  sheep,  and  75  hogs. 
Of  how  many  head  does  his  stock  consist  ? 

Ans,  505  head. 

40.  A  farmer  has  50  acres  planted  in  wheat,  75  in  corn, 
20  in  barley,  14  in  rye,  35  in  oats,  16  in  potatoes,  and  7  in 
"garden  truck."    How  .many  acres  in  all  has  he  planted  ? 

Ans.  217  acres. 

41.  If  in  10  years  the  expenses  of  a  government  are, 
respectively,  $85279524,  $74843279,  $92573181,  $68491183, 
$73111265,  $79400295,  $87765432,  $95556443,  $90200671, 
and  $100785894,  how  much  does  the  government  spend  in 
these  10  years?  Ans.  $848007167. 

42.  In  January  1875,  there  were  31  days;  in  February 
28 ;  in  March  31 ;  in  April  30 ;  in  May  31 ;  in  June  30 ; 
in  July  31 ;  in  August  31  ;  in  September  30  ;  in  October 
31 ;  in  November  30 ;  and  in  December  31.  How  many 
days  were  there  in  the  year  1875  ?  Ans.  365  days. 

43.  By  the  P.,  F.  W.  and  C.  R.  R.  the  distance  from  Pitts- 
burgh to  Alliance  is  83  miles ;  from  Alliance  to  Crestline, 
106  miles ;  from  Crestline  to  Ft.  Wayne,  131  miles  ;  from 
Ft.  Wayne  to  Plymouth,  64  miles ;  from  Plymouth  to  Chi- 
cago, 85  miles.  What  is  the  distance  from  Pittsburgh  to 
Chicago?  A)is.  469  miles. 

44.  Add  24678,  423,  9846,  23,  and  82764. 

45.  Add  4632,  8,  79,  8264,  38000,  and  673289. 

46.  Add  79,  897,  438,  2986,  150,  and  371436. 

47.  Add  873254,  289094,  627380,  and  456987. 

48.  Add  35709,  50804,  90006,  and  47103. 

49.  Add  263,  5827,  39001,  70126,  and  85. 


36  INTEGERS. 

50.  Add  92,  924,  9857,  95064,  and  572. 

51.  Add  8,  28,  348,  4578,  90068,  and  8. 

52.  Add  7,  77,  777,  7777,  7777,  777,  and  77. 

53.  Add  66666,  6666,  666,  66,  6,  66,  666,  and  6666. 
64.  102  +  304  +  506  +  708  +  901  +  206  +  604=:  ? 

55.  2030  +  4050  +  6070  +  8090  + 1230  +  3110  =  ? 

56.  50709  +  2080  +  30604  +  8010  +  9076==:? 

57.  864209  +  753186  +  999999  +  888888  =  ? 

53.  3207  + 10875  +  892  +  6075  + 124607  +  761  =  ? 

59.  456  +  654  +  789  +  897  +  231  + 123  +  875  =  ? 

60.  9  +  99  +  999  +  9999  +  99999  +  999999  +  25  =  ? 

61.  A  merchant  paid  $15796  for  a  lot  and  built  a  house 
upon  it,  which  cost  him  $25650.  How  much  did  the  lot 
and  building  cost  ?  Ans.  $41446. 

62.  A  man  gave  $2375  to  his  son,  $1875  to  his  daughter, 
and  donated  $10984  to  charitable  purposes.  How  much  did 
he  give  away  ?  Ans.  $15234. 

63.  By  selling  a  house  and  lot  for  $25875  I  lost  $987. 
What  did  both  cost  ?  Ans.  $26862. 

64.  A  coal  dealer  shipped  from  Pittsburgh  at  one  time 
225675  bushels  of  coal ;  at  another  time  186798  bushels ; 
at  another  time  75987  bushels.  How  many  bushels  did  he 
ship  in  all  ?  Ans.  488460. 

65.  In  1870  the  population  of  North  America  was 
50,931,242,  South  America  26,677,419,  Central  America 
2,690,635,  and  West  Indies  4,065,945.  What  was  their 
united  population  ?  Ans.  84,365,241. 


OUTLINE   OF   SUBTRACTION. 


TERMS. 


68.  Minuend. 

69.  Subtrahend, 

70.  Difference,  or  JRetnainder, 


o 

M 

H 

< 

H 
PQ 
P 

• 


SIGNS, 


72.  TABLE. 


PRINCIPLES. 


CASES. 


71.  Of  Subtraction, 

59.  Of  Equality. 

60.  Of  Dollars. 


r  73.  ''Same  Kind.** 

74.  ''Same  Order,'' 

76.  "Same  Number  added"  to 

both  Minuend  and  Sub" 
trahend. 

75.  jT. 

77.  ZT. 


78.  RULE. 


79.  PROOF. 


SECTION    IV. 


S;  CBJ  T^R;  AC^'IOjM 


^«S 


67.  Subtraction  is  the  process  of  taking  one  number 
from  another. 

The  Terms  used  in  Subtraction  are  Minuend,  Subtrahend,  and 
Difference,  or  Eemainder. 

68.  The  Minuend  is  that  number  from  which  the 
other  is  to  be  taken. 

69.  The  Subtrahend  is  that  number  which  is  to  be 
taken  from  the  other. 

70.  The  Difference,  or  JRemainder,  is  the  result 
obtained  by  Subtraction. 

71.  The  Sign  of  Subtraction  is  a  short  horizontal 
line  called  minus  (  — ),  and  when  placed  between  two 
numbers  signifies  that  the  one  after  it  is  to  be  taken  from 
the  one  before  it.  Thus,  5  —  2  means  that  2  is  to  be  taken 
from  5,  and  is  read  "  5  minus  2." 

The  term  minus,  Latin,  means  lets,  or  diminished  hy. 

Illustration. — In  the  expression  5—2  =  3,  read  5 
minus  2  equals  3,  5  is  the  rninuend,  2  the  subtrahend,  and 
3  the  difference,  or  remainder. 


SUBTRACTION. 


39 


72.    SUBTRACTION    TABLE 


12    3 
111 

ONE. 

4   5    6    7    8    910 
1111111 

SIX. 

6    7    8    9  10  111213  1415 
66    66666606 

0    12 

3    4   5    6    7    8    9 

0123456    7    89 

2    3    4 
2    2    2 

TWO. 

5-6    7    8    91011 
2    2    2    2    2    2    2 

SEVEN. 

7    8    910111213  1415  16 

7777777777 

0    12 

3    4    5    6    7    8    9 

0123456789 

THREE. 

3    4   5    6    7    8    9101112 
3333333333 

EIGHT. 

8    91011121314151617 
8888888888 

0    12 

3    4   5    6    7    8    9 

0123456789 

4   5    6 
4   4   4 

FOUR. 

7   8    910  111213 
4   4   4   4   4    4   4 

NINE. 

9  10  11 12  13  14  15  16  17  18 
9999999    9    99 

0    12 

3    4   5    6    7    8    9 

0123456789 

5    6    7 
5    5    5 

FIVE. 

8    91011121314 
5    5    5    5    5    5    5 

TEN. 

10111213141516171819 
10  10  10  10  10  10  10  10  10  10 

0    12 

3    4    5    6    7    8    9 

0123456789 

40 


INTEGERS 


0i«al^AfGiici^eX 


Example. — James  had  9  apples  and  gaye  his  brother  4. 
How  many  had  he  left  ? 

SoiiUTiON. — Since  4  apples  from  9  apples  leave  5  apples,,  he  had  5 
apples  left.    Or, 

4  apples  from  9  apples  leave  5  apples.  Therefore,  he  had  5  apples 
left.     Or, 

He  had  5  apples  left,  because  4  apples  from  9  apples  leave  5  apples. 


1.  There  are  10  books  on  the  shelf.  If  I  take  away  8,  how 
many  will  be  left  ?    Take  8  from  10  and  how  many  are  left  ? 

6  from  8?    6  from  8?     1  from  10?    4  from  6?     6  from  7? 

7  from  9?     3  from  10? 

2.  Susan  has  9  cents  and  Mary  has  6.  How  many  more 
has  Susan  than  Mary  ?  How  many  less  has  Mary  than 
Susan.  9  less  6  are  how  many?  9  less  5?  9  less  4? 
9  less  3?    9  less  2?    9  less  1?    9  less  9? 

How  many  are 


3.  15  horses  less  6  horses  ? 

4.  10  sheep  less  4  sheep  ? 

5.  13  bushels  less  7  bushels  ? 

6.  16  miles  less  10  miles? 

7.  12  dollars  less  6  dollars? 

8.  17  cents  less  8  cents? 

9.  14  birds  less  5  birds  ? 

10.  15  marbles  less  8  marbles? 


11.  15  books  less  9  books  ? 

12.  16  pens  less  7  pens  ? 

13.  13  desks  less  6  desks  ? 

14.  5  doors  less  3  doors  ? 

15.  14  chairs  less  8  chairs  ? 

16.  10  boys  less  7  boys  ? 

17.  11  men  less  8  men  ? 

18.  15  figs  less  7  figs? 


19.  If  you  take  7  dollars  from  14  dollars,  how  many 
remain  ? 


SUBTRACTION-. 


41 


20.  Nannie  brought  6  pears  to  school,  and  gave  2  of  them 
to  her  teacher.     How  many  had  she  left  ? 

21.  How  many  are  15  apples  less  8  pears? 

We  cannot  take  8  pears  from  15  apples.     For  15  apples  —  8  pears 
=  neither  7  pears  nor  7  apples.     Hence, 

73.  Only  abstract  numbers,  or  like  concrete  numbers,  can 
be  subtracted. 

22.  From  9  tens  -take  2  ones. 

We  cannot  take  2  ones  from  9  tens.     For  9  tens  —  2  ones  =  neither 
7  tens  nor  7  ones.     Hence, 

74.  Only  like  orders  of  units  can  be  subtracted. 


+ Written  ^^vferei^esT-e 


CASE    I. 

75.  When  each  figure  in  the  Subtrahend  rep- 
resents a  less  value  than  the  correspond- 
ing figure  of  the  Minuend. 

■    Example.— From  68759  subtract  37422. 

SOLUTION.  Explanation.  —  Since   only  like 


Minuend, 
Subtrahend, 

Kemainder, 


68759 
37422 

31337 


orders  can  be  subtracted,  for  conve- 
nience we  write  the  subtrahend  so  that 
its  units  stand  under  the  units  of  the 
same  order  in  the  minuend. 

We  now  subtract  each  figure  of  the 
subtrahend  from  the  figure  above  it  in  the  minuend,  writing  each 
result  in  the  remainder.     2  ones  from  9  ones  leave  7  ones  ;  2  tens  from 

5  tens  leave  8  tens;  4  hundreds  from  7  hundreds  leave  3  hundreds  ; 
7  thousands  from  8  thousands  leave  1  thousand  ;  3  ten-thousands  from 

6  ten-thousands  leave  3  ten-thousands. 

The  result,  31337,  is  the  difference  or  remainder. 


42 


INTEGERS. 


PB.  O  B  LEJUS, 

1.  From  7395  take  2004. 

2.  From  36714  take  12511. 

3.  From  597263  take  84123. 

4.  68764  —  63233  =  ? 

5.  190806  —  90603  =  ? 


Ans.  5391. 

Ans.  24203. 

Ans.  513140. 

6.  68754  —  5521  =  ? 

7.  190856  —  100203  =  ? 

8.  Bought  30  shares  of  bank  stock  for  13789  and  sokl 
them  for  $5999.     How  much  did  I  gain  ? 

9.  A  owns  8  horses  and  B  owns  4.  How  many  more 
has  A  than  B  ?  After  each  has  bought  4  more,  how  many 
more  has  A  than  B  ?    Hence, 

76.  Wlien  the  same  number  is  added  to  both  minuend  and 
subtrahend,  it  does  not  change  their  difference. 


-•>•- 


4-lWi'itten^^Grci^eXt 


m^^^ 

CASE    II. 

77.  Wlien  a  figure  in  the  Subtrahend  represents 
a  greater  value  than  the  corresponding 
figure  of  the  Minuend, 

Example. — From  54  subtract.  19. 


SOLUTION. 

Minuend,  54 

Subtrahend,  19 

Eemainder,  35 


Explanation.— It  is  evident  that  we 
cannot  subtract  9  ones  from  4  ones.  But 
by  adding  10  ones  to  the  minuend  and 
1  ten  (10  ones)  to  the  subtrahend  (76),  we 
can  perform  the  subtraction  very  readily. 

54  =  5  tens  4  ones  ;  add  10  ones  and  we  have        5  tens    14  ones. 
19  =  1  ten  9  ones  ;  add    1  ten    and  we  have        2  tens      9  ones. 
And  2  tens  9  ones  from  5  tens  14  ones  =     3  tens      5  ones, 
or  35,  the  required  difEereuce  or  remainder. 


S  U  B  T  R  A  C  T I  O  K 


43 


Example. - 

SOLUTION. 

924 
376 

548 


-From  924  take  376. 
Explanation. — We  write  the  numbers  as  in  the 
margin  and  say  mentally  : 

6  ones  from  4  ones  impossible ;  but  6  ones  from 
14  ones  leave  8  ones.  Since  we  added  10  ones  (1  ten) 
to  the  minuend,  we  must  add  1  ten  (10  ones)  to  the 
subtrahend.  7  tens  and  1  ten  are  8  tens ;  8  tens  from 
2  tens  impossible ;  but  8  tens  from  12  tens  leave  4  tens.  Since  we 
added  10  tens  (1  hundred)  to  the  minuend,  we  must  add  1  hundred 
(10  tens)  to  the  subtrahend.  3  hundreds  and  1  hundred  are  4  hun- 
dreds ;  4  hundreds  from  9  hundreds  leave  5  hundreds.     Result,  548. 

78.  KuLE. —  Write  the  suUraliend  so  that  its  units  shall 
stand  under  the  units  of  the  same  order  in  the  minuend. 
Begin  at  the  right-hand  and  subtract  from  each  figure  of  the 
minuend  the  corresponding  figure  of  the  subtrahend  and  write 
the  remainder  underneath. 

If  in  the  minuend  any  figure  expresses  a  less  number  of 
ones  than  the  corresponding  figure  of  the  subtrahend ,  add  10 
to  the  former ;  then  subtract,  and  add  1  to  the  next  figure  of 
the  subtrahend.    Proceed  thus  till  the  work  is  complete. 

79.  Proof. —  When  the  work  is  correct,  the  sum  of  the 
remainder  and  subtrahend  equals  the  minuend. 


PI 

lOBLEM^ 

f. 

(1.) 

From         734 

(2.) 

652 

(3.) 

818 

(4.) 
963 

(5.) 
521 

(6.) 
520 

Take          475 

384 

549 

777 

258 

287 

Ans.   259 

From 

233 

7.  403  take  304. 

11. 

900  take  512. 

8.  600  take  259. 

12. 

500  take  201.  ^ 

9.  800  take  468. 

13. 

400  take  102. 

10.  700  take  237. 

14. 

4321  take  1234. 

44  INTEGEKS, 


15.  From  8765  take  5678. 

16.  From  90807  take  6898. 


17.  From  60504  take  496. 

18.  From  30201  take  194. 


19.  How  many  are  100000  —  99999  ?  Ans.  1. 

20.  How  many  are  100000  —  9999?  Ans.  90001. 

21.  How  many  are  100000  —  999  ?  Ans.  99001. 

22.  How  many  are  100000  —  99  ?  Ans.  99901. 

23.  How  many  are  100000  —  9  ?  Ans.  99991. 

24.  By  the  P.  R.  E.  the  distance  from  Pittsburgh  to  New 
York  is  444  miles;  from  Pittsburgh  to  Philadelphia,  355 
miles.  What  is  the  distance  from  Philadelphia  to  New 
York  ?  Ans.  89  miles. 

25.  By  the  same  road,  from  Pittsburgh  to  Harrisburg  is 
246  miles.  What  is  the  distance  of  Harrisburg  from  Phila- 
delphia?   From  New  York?  1st  Ans.  109  miles. 

2d  Ans.  198  miles. 

26.  Washington  was  born  A.  d.  1732,  and  died'in  A.  d.  1799. 
How  old  was  he  at  his  death  ?  Ans.  611  years. 

27.  The  sum  of  two  numbers  is  20560 ;  the  less  is  3745. 
What  is  the  greater?  Ans.  16815. 

28.  A  merchant  sold  on  Saturday  goods  amounting  to 
$2567  ;  on  Monday  goods  amounting  to  $1075.  What  is  the 
difference  in  the  two  days'  sales  ?  Ans.  $1492. 

29.  If  a  man's  property  is  valued  at  $15725,  and  his  debts 
at  $6837,  what  is  he  worth  ?  Ans.  88888. 

30.  If  a  man's  property  is  valued  at  $20569,  and  his  debts 
at  $30880,  what  excess  of  debt  has  he  ?  Aiis.  $10311. 

31.  A  man  sold  a  horse  for  $375,  making  $97.  How 
much  did  the  horse  cost  him  ?  Ans.  $278. 

32.  A  and  B  together  bought  property  for  $275315,  and 
A  furnished  $98676  of  the  purchase  money.  How  much 
did  B  furnish  ?  Ans.  $176639. 


SUBTRACTION.  45 

33.  America  was  discovered  by  Columbus  in  1492.  How 
many  years  had  elapsed  in  1837  ?  Ans.  345  years. 

34.  I  deposited  in  the  bank  $1840,  and  drew  out  M75. 
How  many  dollars  had  I  left  ?  Ans,  $1365. 

35.  A  man  has  property  worth  $10104,  and  owes  debts  to 
the  amount  of  $7426.  When  his  debts  are  paid,  how  much 
will  he  have  left  ?  Ans.  $2678. 

36.  A  man  having  $100000,  gave  away  $11.  How  many 
dollars  had  he  left  ?  Ans.  $99989. 

37.  A  man  bought  a  span  of  horses  for  $537  and  a  yoke 
of  oxen  for  $297.  How  much  more  did  he  pay  for  the 
horses  than  for  the  oxen?  Ans.  $240. 

38.  A  man  having  $8795  gave  $2309  of  it  for  a  store. 
How  much  money  had  he  left?  Ans.  $6486. 

39.  A  merchant  owing  $35542  paid  $22560.  How  much 
does  he  still  owe  ?  Ans.  $12982. 

40.  A  merchant  selling  goods  at  $6789,  gained  $444. 
Whatdid  they  cost?  Ans.  $6345. 

41.  Chatham  is  47  miles  west  of  York,  and  Howe  is 
91  miles  west  of  York.  How  far  is  it  from  Howe  to 
Chatham  ?  Ans.  4A  miles. 

42.  If  the  distance  of  the  moon  from  us  is  240000  miles, 
and  that  of  the  sun  95000000  miles,  how  much  farther  is  it 
to  the  sun  than  to  the  moon  ?  Ans.  94760000  miles. 

43.  Washington  died  A.  d.  1799,  at  the  age  of  67.  In 
what  year  was  he  born  ?  Ans.  1732. 

44.  At  an  election  the  successful  candidate  received  5075 
votes,  and  had  a  majority  of  387  over  the  defeated  candi- 
date.    How  many  votes  did  the  latter  receive  ? 

Ans.  4688  votes. 


46  IKTEGERS. 

45.  How  many  years  is  it  since  the  discovery  of  America 
by  Columbus  in  1492  ? 

46.  Find  the  difference  between  eight  thousand  seventeen, 
and  seven  thousand  eighteen. 

47.  An  army  numbered  before  a  battle  thirty  thousand 
five  hundred  ten  men ;  after  battle  twenty-six  thousand 
seven  hundred  nineteen.    What  was  the  loss  ? 

Ans.  3791  men. 

Find  the  difference 

48.  Between  85374  and  5805. 

49.  Between  4709  and  G3489. 

50.  Between  75310  and  864. 

51.  Between  9811  and  98021. 

52.  Between  5678901  and  991112. 

53.  Between  82223  and  1359012. 

54.  Between  26590123  and  733334. 

55.  Between  6444445  and  924901123. 

56.  4159012345  —  86665554  =  ? 

57.  65490123456  —  566666667  =  ? 

58.  The  difference  of  two  numbers  is  753,  and  the  greater 
is  1031.     What  is  the  less  ?  Ans,  278. 

59.  What  number  added  to  2056  makes  18000  ? 

60.  What  number  taken  from  67890  leaves  eGGQH  ? 

61.  The  thermometer  stood  at  19  degrees  above  zero  in 
the  morning,  and  45  degrees  above  zero  in  the  afternoon. 
How  many  degrees  did  it  rise  ? 

62.  A  farmer  raised  1220  bushels  of  wheat,  and  700  bushels 
of  com.  He  sold  875  bushels  of  wheat,  and  225  bushels  of 
corn.     How  many  bushels  of  each  had  he  left  ? 

Ans.  345  of  wheat ;  475,  corn. 


OUTLINE   OF   MULTIPLICATION. 

81.  Product, 

83.  Multiplicand. 


TERMS. 


82.  Factors. 


(84. 


Multiplier. 


r  85.  Of  Mtatiplication. 

• 

o 

SIGNS. 

69.   Of  Equality, 
60.   Of  Dollars, 

M 

H 

.  95\  Parenthesis, —Vinculum 

< 

M 
1-^ 

86.  TABLE. 

H 

'  87.  J^irs*. 

H 

88.  Second, 

PRINCIPLES.   - 

89.  T/wrcl. 

90.  Fourth, 

g 

6 

X 

r  91.  First. 

CASES. 

92.  Second. 
I  94.  T/*ir<i. 

RULES. 


93.  l?V>r  Case  JJ. 
95.  -For  Case  2JJ. 


SECTION    V. 


-^g — g^e^p-^aer- 


m  mmiM'i'BiMi€ATi&>m  I 


^- 


^,%<^^-^ 


80.  Multiplication  is  the  process  of  repeating  one 
riumber  as  many  times  as  there  are  ones  in  another. 

Or,  it  is  a  short  process  of  adding  equal  numbers. 
The  Terms  are  Prodicct  and  Factors. 

81.  The  Product  is  the  result  obtained  by  Multipli- 
cation. 

82.  The  Factors  are  the  numbers  used  to  obtain  the 
Product. 

"YYie  factors  are  Multiplicand  and  Multiplier. 

83.  The  Multiplicand  is  that  factor  which  is  to  be 
repeated. 

84.  The  Multiplier  is  that  factor  which  shows  how 

many  times  the  Multiplicand  is  to  be  repeated. 

When  it  is  said,  2  times  3  are  6,  6  is  the  Product  and  3  and  3  are 
the  Factors,  of  which  2  is  the  Multiplier  and  3  the  Multiplicand. 

85.  The  Sign  of  Multiplication  is  an  oblique 
cross  (  X ),  and  when  placed  between  two  or  more  numbers 
signifies  that  they  are  to  be  multiplied  together.  Thus, 
2x3  means  that  2  and  3  are  to  be  multiplied  together,  and 
is  read  "2  times  3." 


MULTIPLICATION. 


49 


86.    MULTIPLICATION    TABLE 


0    1 

1  1 

ONE. 

234567    8    9 
11111111 

0 
6 

SIX. 

123456789 
66666    6-666 

0    1 

23456789 

0 

6  12  18  24  30  36  42  48  54 

0    1 

2    2 

TWO. 

2345    6    789 
22222222 

0 

7 

SEVEN. 

123456789 

777777777 

0   2 

4    6    810  12141618 

0 

7  14  21  28  35  42  49  56  63 

0    1 
3    3 

THREE. 

23    456789 
33333333 

0 

8 

EIGHT. 

123456789 

888888888 

0    3 

6    9121518  2124  27 

0 

8  16  24  32  40  48  56  64  72 

0   1 
4   4 

0   4 

FOUR. 

23456789 
4   4_4_4_4_4jt_4 

8  12  16  20  24  28  32  36 

0 
9 

0 

NINE. 

123456789 
999999999 

9  18  27  36  45  54  63  72  81 

0    1 
5    5 

0    5 

FIVE. 

23456789 
55555555 

TEN. 

0123456789 
10  10  10  10  10  10  10  10  10  10 

10  15  20  25  30  35  40  45 

0  10  20  30  40  50  60  70  80  90 

50 


INTEGERS. 


..-f.- 


-f  ©ral^ 


Example. — What  will  4  apples  cost  at  3  cents  each  ? 

Solution.— Since  4  times  3  cents  are  13  cents,  4  apples  at  3  cents 
each  will  cost  12  cents.    Or, 

4  times  3  cents  are  12  cents.  Therefore,  4  apples. at  3  cents  each 
will  cost  12  cents.    Or, 

4  apples  at  3  cents  each  will  cost  12  cents,  because  4  times  3  cents 
are  12  cents.    Or, 

If  1  apple  cost  3  cents,  4  apples  will  cost  4  times  3  cents  or  12  cents. 
Therefore,  4  apples  at  3  cents  each  will  cost  12  cents. 


PBOBI^EMS 

What  cost 

1.  9  pencils  at  6  cents  each  ? 

2.  7  chairs  at  8  dollars  each  ? 

3.  5  hats  at  4  dollars  each  ? 

4.  3  lemons  at  5  cents  each  ? 

5.  4  tons  of  coal  at  $9  a  ton  ? 

6.  8  harrels  of  flour  at  $8  a  barrel  ? 

7.  6  yards  of  cloth  at  $7  a  yard  ? 

8.  6  vests  at  5  dollars  each  ? 

9.  9  books  at  3  dollars  each  ? 


How  many  are 

11.  9x4cents?  4x9? 

12.  7x5dollars?  5x7? 

13.  5x9dollars?  9x5? 

14.  3  X  8  cents  ?  8  x  3  ? 

15.  4x8dollars?  8x4? 

16.  8x9dollars?  9x8? 

17.  6x8  cents?  8x6? 

18.  6x3dollars?  3x6? 

19.  9x7dollars?  7x9? 

20.  7x3dollars?  3x7? 


10.  7  pairs  of  shoes  at  $4  a  pair  ? 

21.  If  8  pounds  of  beef  will  last  a  family  one  week,  how 
many  pounds  will  last  them  5  weeks  ?   2  weeks  ?    7  weeks  ? 

Example. — How  many  are  4  chairs  x  3  dollars  ? 

We  cannot  multiply  3  dollars  by  4  chairs.  For  the  product  would 
neither  be  12  chairs,  nor  12  dollars. 

We  can  multiply  3  dollars  by  4.  For  4  times  3  dollars  are 
($3  +  $3  + 13  +  $3),  or  12  dollars.    Hence, 


MULTIPLICATION^. 


51 


87.  The  Multiplier  is  always  an  abstract  number, 

88.  The  Multiplicand  may  be  an  abstract,  or  concrete 
number. 

89.  The  Product  is  always  of  the  same  kind  as  the  true 

Multiplicand. 

Note. — In  the  explanation  of  the  solution  of  problems  which  contain 
concrete  numbers,  the  number  used  concretely  is  the  true  multiplicand. 
Thus,  4  chairs  at  |3  each,  will  cost  4  times  "3  dollars"  etc.  In  this 
explanation  "  3  dollars"  is  the  number  used  concretely,  and  is,  there- 
fore, the  true  multiplicand. 

90.  The  product  is  numerically  the  same  in  whatever 
order  the  factors  are  multiplied  together. 


i'lWi«itten^xferci^eX'+ 


CASE    I. 
91.  When  the  Multiplier  does  not  exceed  9, 

Example. — How  many  are  564-f  564+564,  or  3  x  564  ? 

Explanation.— In  the  Solution  by  Addition 
we  find  that  the  sum  of  3  564's  is  1692.  But 
since  4  ones  +  4  ones  +  4  ones  =  3x4  ones,  and 
6  tens  +  6  tens  +  6  tens  =  3x6  tens,  and 
5  hundreds  +  5  hundreds  +  5  hundreds  =  3  x 
5  hundreds,  the  process  can  be  very  much 
shortened,  as  shown  in  the  following  Solution 
by  Multiplication. 

Explanation.— In  this  Solution  the 
Multiplicand  564  is  written  but  once  ; 
and  as  it  is  to  be  repeated  3  times,  we 
write  the  Multiplier  3  under  it,  and 
commence  at  the  right  hand  to  mul. 
tiply. 


solution. 
By  Addition. 

i    564 

Parts,  -j    564 

(    564 

Sum,  1692 


solution. 

By  Multiplication. 

Multiplicand,    564 
Multiplier,      3 

Product,  1692 


O'Z  IKTEGERS. 

3  times  4  ones  are  12  ones  =  1  ten  and  2  ones.  Write  the  2  ones  in 
ones'  order  of  the  product  and  reserve  the  1  ten  to  add  to  the  product 
of  tens.  3  times  6  tens  are  18  tens,  plus  1  ten  reserved,  are  19  tens 
=  1  hundred  and  9  tens.  Write  the  9  tens  in  tens'  order,  and  reserve 
the  1  hundred  to  add  to  the  product  of  hundreds.  3  times  5  hundreds 
are  fifteen  hundreds,  plus  1  hundred  reserved,  are  16  hundreds 
=  1  thousand  and  6  hundreds,  which  write  in  the  thousands'  and  hun- 
dreds' orders. 

Hence  the  Product  is  1692.  equal  to  the  Sum  of  the  1st  Solution. 

PROBLEMS, 

Solve  by  both  methods. 

1.  4  times  83.  4.  6  times  344.  7.  2  times  476. 

2.  5  times  126.        5.  7  times  519.  8.  8  times  155. 

3.  3  times  693.        6.  4  times  195.  9.  9  times  619. 

10.  Multiply  372  by  3  ;    by  4 ;    by  5 ;    by  6  ;    by  7. 

11.  Multiply  5344  by  4 ;    by  6 ;    by  8 ;     by  9 ;    by  5. 

12.  Multiply  474372  by  5  ;    by  7 ;     by  9 ;    by  6. 


13.  67391  X  6  =  ? 

14.  36519  X  4  =  ? 

15.  62168  x7  =  ? 


16.  537354  x  9  =  ? 

17.  313763  X  8  =  ? 

18.  898767  x  5  =  ? 

19.  What  cost  456  barrels  of  flour,  at  $8  a  barrel? 

Although  $8  is  the  true  Multiplicand,  for  convenience  we  use  8  for 
the  Multiplier  and  456  for  the  Multiplicand  (SO,  Note),  but  the  Pro- 
duct is  dollars.  For  if  1  barrel  of  flour  cost  $8,  456  barrels  will  cost 
456  times  $8  ==  $3643. 

20.  What  cost  986  pounds  of  sugar,  at  9  cents  a  pound  ? 

21.  What  is  the  cost  of  7  lots  of  ground,  at  15648  each  ? 

22.  How  much  will  8  tons  of  hay  cost,  at  $23  a  ton  ? 

Ans.  $184. 

23.  How  much  will  4  horses  cost,  at  1225  each  ? 

Ans.  $900. 

24.  How  much  will  a  year's  board  cost,  at  $6  a  week,  there 
being  52  weeks  in  a  year  ?  Ans.  $312. 


MULTIPLICATION^.  63 

25.  There  are  5280  feet  in  a  mile.  How  many  feet  in 
9  miles?  Ans.  47520  feet. 

26.  There  are  1760  yards  in  a  mile.  How  many  yards  in 
7  miles  ?  A7is.  12320  yards. 

27.  How  far  will  a  train  of  cars  run  in  8  hours,  at  the 
rate  of  42  miles  an  hour  ?  Ans.  336  miles. 

28.  An  iron  manufacturer  pays  his  hands  $4269  a  month. 
How  much  do  their  wages  amount  to  in  6  months  ? 
in  7  months?  1st  Ans.  $25614  ;  2d,  $29883. 

29.  What  will  8  farms  cost  at  $12219  each? 

30.  How  many  square  miles  in  7  townships,  if  each  con- 
tains 24375  square  miles  ? 

31.  How  much  will  328  yards  of  cloth  cost,  at  $5  per 
yard?  Ans.  $1640. 

32.  What  will  375  head  of  sheep  cost,  at  4  dollars  a  head  ? 

Ans.  $1500. 

33.  What  will  2768  barrels  of  flour  cost,  at  $7  a  barrel? 

Ans.  $19376. 

34.  There  are  1728  cubic  inches  in  one  cubic  foot.  How 
many  cubic  inches  in  8  cubic  feet  ? 

Ans.  13824  cubic  inches. 

35.  In  one  square  foot  there  are  144  square  inches.  How 
many  square  inches  in  6  square  feet. 

Ans.  864  square  inches. 

36.  At  $325  per  acre  what  will  a  lot  of  9  acres  cost  ? 

Ans.  $2925. 

37.  What  cost  675  melons  at  8  cents  each  ? 

38.  What  will  7  buggies  cost  at  $225  apiece  ? 

39.  A  man  travels  8  miles  an  hour.  How  far  can  he 
travel  in  4897  hours?  Ans.  39176  miles. 


54 


INTEGERS, 


^-iVTritten^xfercisesr-f 


CASE    11^ 
93.  When  the  Multiplier  exceeds  9. 

Example.— Multiply  583  by  47. 


583 
47 


SOLUTION. 

Multiplicand, 
Multiplier, 

1st  Partial  Product,     4081 
2d    Partial  Product,  2332 

Product,  27401 


Explanation. — We    first 

multiply  583  by  7,  the  num- 
ber of  ones  in  the  Multiplier 
as  taught  in  Case  I,  and 
obtain  4081. 

We  then  in  the  same  man- 
ner multiply  583  by  4,  the 
number  of  tens  in  the  Multi- 
plier and  obtain  2332. 
But  since  1  ten  is  10  times  as  great  as  1  one,  4  tens  must  be 
10  times  as  great  as  4  ones,  and  therefore  each  figure  multiplied  by 
4  tens  will  produce  a  product  ten  times  as  great  as  the  same  figure 
multiplied  by  4  ones.  Therefore  583  multiplied  by  4  tens,  is  ten 
times  as  great  as  when  multiplied  by  4  ones. 

But  when  multiplied  by  4  ones  the  product  is  written  so  that  the 
right-hand  figure  stands  in  the  ones'  place.  Hence,  to  make  it  express 
ten  times  as  much,  it  must  be  written  so  that  the  right-hand  figure 
shall  stand  m  the  tens'  place. 

Writing  it  thus  and  adding  the  two  Partial  Products,  we  have 
27401,  the  required  Product. 


93.  EuLE. —  Write  the  Multiplier  under  the  Multiplicand 
so  that  units  of  the  same  order  shall  stand  in  the  same 
column. 

Multiply  the  entire  Multiplicand  hy  each  significant 
figure  in  the  Multiplier,  and  write  the  several  Partial 
Products  ona  under  another  so  that  the  right-hand  figure  of 


MULTIPLICATION^.  55 

each  Partial  Product  will  represent  the  same  order  as  the 
figure  of  the  Multiplier  producing  it. 

The  sum  of  the  Partial  Products  is  the  required  Product. 


1.  364805  X  92  =  ? 

2.  213254  X  607  :=  ? 

3.  380097  X  504  =  ? 

4.  400562  X  904  =  ? 

5.  576348  x  746  =  ? 


JPiJ  O  B  LEMS. 

6.  908070  x   908  =:  ? 

7.  174563  X 1385  =  ? 

8.  217572x4175  =  ? 

9.  764050  X 1731  =  ? 
10.  374781  X 1402  =  ? 


These  10  Problems  may  be  increased  in  number  to  100  by  multi- 
plying eacb  Multiplicand  by  eacli  of  the  10  given  Multipliers. 

11.  If  a  vessel  sails  231  miles  in  1  day,  how  far  will  she 
sail  in  35  days  ?  in  72  days  ?  1st  Ans.  8085  miles. 

12.  If  a  man  spends  $25  a  day  for  store  expenses,  how 
much  will  he  spend  in  730  days?  Ans.  $18250. 

13.  A  man  bought  16  lots  at  1750  apiece.  How  much 
did  they  all  cost  him  ?  Ans.  $12000. 

14.  A  farmer  raised  456  bushels  of  corn  each  year  for 
19  years.     How  much  in  all  did  he  raise  ? 

Ans.  8664  bushels. 

15.  A  gentleman's  expenses  for  1  year  were  $3450.  At 
the  same  rate  what  would  they  be  for  25  years  ? 

Ans.  $86250. 

16.  In  a  school  there  are  19  teachers  and  43  pupils  for 
each  teacher.    How  many  pupils  are  there  ? 

Ans.  817  pupils. 

17.  In  a  certain  school  district  there  are  351  pupils,  and 
the  cost  of  tuition  per  pupil  is  $14.  What  is  the  entire 
cost?"  Ans.  $4914. 

18.  What  cost  450  tons  of  iron,  at  $45  per  ton  ? 


66 


INTEGEES. 


19.  What  cost  37  tons  of  steel,  at  $63  per  ton  ? 

20.  What  cost  372  yards  of  cloth,  at  $9  per  yard  ? 

21.  What  cost  568  acres  of  ground,  at  $175  per  acre  ? 

22.  A  railroad  company  bought  13  locomotives  at  19500 
apiece.    What  did  they  all  cost  ?  Ans.  $123500. 

23.  An  institution  spends  in  advertising  $565  per  year; 
at  this  rate  how  much  would  it  spend  in  15  years  ? 

Ans.  $8475. 

24.  Multiply  three  hundred  forty-one  million  six  hundred 
four  thousand  five  hundred  one  by  twenty-three  thousand 
six  hundred  eight.  Ans.  8064599059608. 

25.  Find  the  product  of  seventy-five  million  three  hun- 
dred sixty-seven  thousand  four  hundred  fifty-one  by  seven 
thousand  twenty-one.  Ans.  529154873471. 


4^  Wi  itten  ^JExTerei^eXt 


•'•^ — - 

CASE    III. 

94.  When  either  factor  has  one  or  more  ci" 
phers  on  the  right. 

Example.— Multiply  325  by  10 ;  by  100. 

Explanation. — Since  10  units  of 
any  order  make  1  of  the  next  higher 
order,  the  writing  of  a  cipher  on  the 
right  of  a  number,  thus  removing  each 
of  its  figures  one  order  to  the  left,  must 
multiply  it  by  10.     In  like  manner 

the  writing  of  two  ciphers  on  the  right  of  a  number  multiplies  it 

by  too. 

Hence,  to  325  x  1,  which  is  325,  we  annex  one  cipher,  and  have  10 

times  the  number;  annex  two  ciphers  and  we  have  100  times  the 

number. 


SOLUTION. 

Multiplicand,  325 
Multiplier,  10 

Product,  3250 


MULTIPLICATION.  67 

Example. — Multiply  325  by  50  ;  by  500. 

SOLUTION.  Explanation.— Since  50  is  10  times  5,  50  times 

325  325  must  be  10  times  5  times  325. 

-^  5  times  325  is  1625,  and  10  times  as  much,  is  1625 

with  one  cipher  annexed,  or  16250. 

16250  In  like  manner  the  annexing  of  two  ciphers  gives 

the  product  when  325  is  multiplied  by  500. 

Example.— Multiply  3250  by  50. 

solution.  Explanation.— 5  times  3250  is  5  times  325  with  a 

3250  cipher  annexed,    or  16250,    and    50  times    3250    is 

KQ  16250  with  one  cipher  annexed,  or  162500,  the  re- 

quired  product. 

162500 

95.  Rule. — To  the  product  of  all  the  other  figures  annex 
as  many  ciphers  as  there  are  at  the  right  of  both  factors. 

PR  OBZE  MS. 

1.  Multiply  375  by  100  ;  by  420:  41900  by  9090. 

2.  Multiply  1450  by  3500;  by  780:  3080  by  1450. 

3.  Multiply  6300  by  1700;  by  4000:  7500  by  6300. 

4.  Multiply  245  by  36000;  by  790:  1260  by  2450. 
.  5.  Multiply  375000  by  360  ;  by  42 :  1420  by  3750. 

6.  Multiply  5090  by  480;  by  90  :  6030  by  5090. 

7.  Multiply  6203  by  9300;  by  200:  4090  by  6200. 

8.  A  barrel  of  flour  weighs   196   pounds.    How  many 
pounds  in  200  barrels  ?  Ans.  39200  pounds. 

9.  A  barrel  of  fish  weighs  200  pounds.      How  many 
pounds  in  150  barrels  ?  Ans.  30000  pounds. 

10.  A  barrel  of  salt  weighs  280   pounds.     How  many 
pounds  in  179  barrels  ?  Ans.  50120  pounds. 

11.  A  barrel  of  lime  weighs  240  pounds.     How  many 
pounds  in  56  barrels  ?  Ans.  13440  pounds. 


68  INTEGERS. 

12.  A   bushel  of  coal   weighs   76  pounds.    How  many 
pounds  in  4800  bushels  ?  Ans.  364800  pounds. 

13.  How  much  will  20  city  lots  cost,  at  14100  each  ? 

Ans.  $82000. 

14.  How  much  will  16  acres  of  ground  cost,  at  $170  per 
acre?  Ans.  12720. 

15.  How  much  will  450  head  of  cattle  cost,  at  $56  a  head  ? 

Ans.  $25200. 

16.  How  much  will  351  head  of  sheep  cost,  at  $10  a  head  ? 

Ans.  $3510. 

17.  How  much  will  402  head  of  horses  cost,  at  $230  a 
head?  A7is.  $92460. 

18.  In  printing  5000  copies  of  a  book,  24  sheets  of  paper 
were  used  for  each  book.     How  much  paper  was  used  ? 

Ans.  120000  sheets. 

19.  In  building  a  line  of  telegraph  370  miles  long,  the 
cost  was  $1004  per  mile.    What  was  the  entire  cost  ? 

Ans,  $371480. 

20.  At  a  plow  factory  108  plows  were  made  each  day  for 
50  days.     How  many  plows  were  made  ?   Ans.  5400  plows. 

21.  An  agent  travels  297  days  at  an  average  of  125  miles 
miles  per  day.    Howmany  miles  does  he  travel  in  the  year  ? 

Ans.  37125  miles. 

22.  How  many  days  in  8  years  of  365  days  each  ? 

Ans.  2920  days. 

95'.  A  Parenthesis  (  )  is  used  to  include  such  num- 
bers as  are  to  be  considered  together. 
A  Vinculum  has  the  same  meaning. 


Thus,  4  X  (5  +  4),  or  4  X  5  +  4  means  that  4  is  to  be  multiplied  by 
5  +  4  or  9.     4  multiplied  by  9  =  36,  Ans. 

Without  the  parenthesis,  or  vinculum,  4x5  +  4  means  4  times  5  and 
4  added.     4  times  5  =  20 ;  20  +  4  =  24,  Ans. 


MULTIPLICATION.  69 


hlWi«itten?EXfepci*e^-f 


REVIEW    PROBLEMS. 

1.  Multiply  325  —  (15  x  15)  by  47  x  (1861  —  1814). 

Ans.  220900. 

2.  Multiply  801  —  169  +  533  by  624  —  25~^2r. 

Ans.  9801. 

3.  Multiply  (40  x  50)  — 1850  by  (91  x  19)  — 1579. 

Ans.  22500. 

4.  Multiply  (75  x  24)  —  849  by  (63  x  38)  -  1393. 

Ans.  951951. 

5.  Multiply  42  +  (13  x  43)  by  1761  -  (38  x  23). 

Ans.  533087. 

6.  75  +  8  X  (25  +  38)  x  (89  —  72)  x  7  —  100  =  ? 

Alls.  59951. 

7.  20  X  (16  X  5  -  14  X  3)  +  (42  -  18  x  2)  -  700  rzz  ? 

Ans.  QQ, 

8.  15291  —  1342  +  863  —  14812  =  ? 

9.  1764  +  91426  -  92190  +  HI  =  ? 

10.  What  sum  must  be  added  to  271  x  300  to  make  the 
amount  131313  ?  Ans.  50013. 

11.  What  is  the  difference  between  75011  —  (671  x  102) 
and  463  x  201  —  86394  ? 

12.  (463  +  537—901)  x  990  =  ?  Ans.  98010. 

13.  96  +  8  X  (35  +  40)  by  (56— 40x8)— 5  =  ? 

Ans.  85608. 

14.  A  speculator  bought  800  acres  of  land  at  $20  an  acre ; 
he  sold  at  one  time  350  acres  at  $25  an  acre ;  at  another 
time  250  acres  at  $30  an  acre.  At  what  price  must  he  sell 
the  remainder  in  order  to  gain  on  purchase  $8250. 

Ans.  $8000. 


OUTLINE   FOR   DIVISION. 


O 

M 
H 

H 


9^ 


TERMS. 


SIGNS. 


97.  Dividend,  (Prodiict.) 

98.  Divisor,      (Factor.) 

99.  Quotient,    {Factor.) 

100.  O/  Division. 

59.  O/  Equality. 

60.  O/  Dollars. 

95'.  Parenthesis.—  Vinculum, 


101.  TABLE. 

102.  OBJECT,  OR  OFFICE. 


PRINCIPLES. 


103.  l^insf. 

104.  Second. 


CASES. 


r  105.  J. 

107.  II. 
L  109.  JJJ. 


RULES. 


106.  Case  I. 

108.  Case  II. 

L  no.  Case  III. 


SECTION   VIo 

• g)  ®  (5 » 


^ 


miWISiIO;M 


96.  Division  is  the  process  of  finding  how  many 
times  one  number  is  contained  in  another. 

The  Tertns  in  Division  are  Dividend,  Divisor,  and  Quotient. 

97.  The  Dividend  is  the  number  to  be  divided. 

98.  The  Divisor  is  the  number  by  which  to  divide. 

99.  The  Quotient  is  the  result  obtained  by  Division. 

1.  When  there  is  nothing  left  after  division,  the  division  is  said  to 
be  exact ;  and  the  divisor  is  called  an  exact  divisor. 

2.  That  part  of  the  dividend  which  is  left  when  the  division  is  not 
exact  is  called  the  Remainder. 

3   The  remainder  must  always  be  of  the  same  denomination  as  the 
dividend,  because  it  is  really  a  part  of  it. 
4.  A  true  remainder  is  always  less  than  the  divisor. 

100.  The  Sign  of  Division  is  a  short  horizontal 

line  placed  between  two  dots  (-^),  and  when  placed  between 

two   numbers   signifies  that  the  one   on  the  left  is  to  be 

divided  by  the  one  on  the  right.     Thus,  Q  -\-2  means  that 

6  is  to  be  divided  by  2,  and  is  read  "  6  divided  by  2." 

Division  is  sometimes  indicated  by  writinor  the  dividend  above,  and 
the  divisor  below  a  short  horizontal  line.  Thus,  f  is  read  "  6  divided 
by  2."  In  writinqr  numbers  for  solution,  the  Divisor  and  Dividend 
may  be  separated  by  a  curved  line.  Thus,  2  )  C,  or  6  ( 2  is  read  "  G  divid- 
ed by  2."  "      ~ 


6X 


62 


INTEGERS. 


101.    DIVISION    TABLE. 


ONE. 
1)1    23456789 
123456789 

SEVEN. 

7  V?  14  21  28  35  42  49  56  63 
12    3    4   5    6    7~8~9 

TWO. 
2)^    4   6    81012141618 
123    4   56789 

EIGHT. 
8)81624  32  40  48  56  64  72 
123456789 

THREE. 
3)3    6    91215  18  212427 

NINE. 
9  )  9  18  27  36  45  54  63  72  81 

123456789 

123456789 

FOUR. 
4)4   81216  20  24  28  32  36 

TEN. 
10)10  20  30  40  50  60  70  80  90 

123456789 

123456789 

FIVE, 
5)5  1015  20  2530  35  40  45 

ELEVEN. 
11)11  22  33  44  55  66  77  88  99 

123456789 

123456789 

SIX. 
6)61218  24  30  36  42  48  54 

T  W  EL  VE. 
12)12  24  36  48  60  72  84  96108 

123456789 

12345678      9 

DIVISION.  63 

10^.  The  Object  of  Division  is  twofold: 

First  To  find  how  many  times  one  number  is  contained 
in  another. 

Second.  To  find  one  of  the  equal  parts  of  a  number. 

First.  Example. — A  man  has  10  dollars  with  which  he 
wishes  to  buy  books  at  2  dollars  each.  How  many  books 
can  he  buy  ? 

Solution. — Since  2  dollars  a/re  contained  in  10  dollars  5  times,  he  can 
buy  5  hooks. 

Or,  2  dollars  are  contained  in  10  dollars  5  times.  Therefore,  he  can 
buy  5  looks. 

Second.  Example. — If  a  man  divides  10  dollars  equally 
between  2  persons,  how  much  does  each  receive  ? 

Solution. — Since  2  persons  receive  10  dollars,  one  person  receives 
ONE-HALF  of  10  dollars,  or  5  dollars. 

Or,  One-half  of  10  dollars  is  5  dollars.  Therefore,  each  person 
receives  5  dollars. 

The  preceding  are  representative  examples  of  the  two  kinds  of 
problems  to  which  Division  is  applied. 

In  the  fird,  the  Divisor  and  Dividend  are  like  numbers  and  the 
Quotient  is  an  abstract  number.    Hence, 

103.  The  Quotient  is  an  abstract  number,  when  the 
Divisor  and  Dividend  are  both  abstract,  or  both  concrete 
numbers. 

In  the  second  example,  the  Divisor  and  Dividend  are  unlike  numbers 
and  the  Quotient  is  the  same  kind  as  the  dividend.    Hence, 

104.  When  the  Divisor  and  Dividend  are  unlike  num- 
bers the  Quotient  is  alivays  the  same  kind  as  the  Dividend. 

The  mode  of  reasoning  in  the  solution  of  these  two  classes  of  exam- 
ples is  somewhat  different ;  but  in  reality  the  operation  is  the  same, 
viz.:  to  find  how  many  times  one  number  is  contained  in  another, 
which  is  in  harmony  with  the  definition  of  Division  (96). 


64  INTEGERS 


0i»at  iExfGiicis%X  -i 


^^ How  Many  Tinies,'^ 

1.  How  many  pounds  of  rice,  at  8  cents  a  pound,  can  be 
bought  for  56  cents  ?     for  24  cents  ? 

2.  When  raisins  are  9  cents  a  pound,  how  many  pounds 
can  be  purchased  for  72  cents  ?    for  36  cents  ? 

3.  At  8  cents  a  bushel,  how  many  bushels  of  coal  can  be 
bought  for  64  cents  ?     for  48  cents  ? 

4.  At  $7  per  ton,  how  much  ice  can  be  bought  for  $63  ? 
for  135  ?    for  149  ? 

5.  How  many  pencils,  at  6  cents  each,  can  be  bought  for 
48  cents  ?     for  54  cents  ? 

6.  To  how  many  boys  can  I  give  $30,  giving  each  boy  $5  ? 
giving  each  $6  ? 

7.  At  4  cents  each,  how  many  oranges  can  I  buy  for 
36  cents  ?     for  16  cents  ? 

8.  At  6  cents  a  quart,  how  many  quarts  of  milk  can  be 
bought  for  24  cents  ?    for  36  cents  ? 

9.  A  man  paid  81  cents  for  beef,  at  9  cents  a  pound. 
How  many  pounds  did  he  buy  ? 

10.  A  lady  gave  16  apples  to  some  boys,  giving  4  apples 
to  each  boy.    How  many  boys  were  there  ? 

11.  How  many  times  can  a  bucket  which  holds  3  gallons 
be  filled  from  a  barrel  which  holds  27  gallons  ? 

12.  To  how  many  persons  can  I  give  $14,  giving  each 
person  $2  ? 

13.  At  7  cents  each  how  many  oranges  can  I  buy  for 
63  cents  ? 


Pivisioiiir.  65 


©i*al  ^.^or  cisG^ 


''  Equal  Parts.'^ 

When  a  number  or  tiling  is  divided  into  two  equal  parts,  the  parts 
are  called  halves ;  into  three,  the  parts  are  called  thirds ;  into  four, 
fourths;  into  Jive,  fifths;  into  ten,  tenths,  etc. 

A  number  is  divided  into  two,  three,  four,  five,  ten,  etc.,  equal  parts 
^y  dividing  by  2,  3,  4,  5,  10,  etc.,  respectively. 

1.  What  is  a  half  of  8  ?     Of  10  ?    Of  18?    Of  12? 

2.  What  is  a  third  of  9  ?     Of  12  ?     Of  24  ?    Of  6  ? 

3.  What  is  a  fourth  of  10  ?     Of  8  ?    Of  36  ?     Of  12  ? 

4.  What  is  a  fifth  of  20  ?  •  Of  45  ?    Of  20?     Of  35  ? 

5.  What  is  a  sixth  of  18  ?    Of  54?    Of  24  ?  Of  36  ? 

6.  What  is  a  seventh  of  14  ?     Of  28  ?    Of  63  ?     Of  35  ? 

7.  What  is  an  eigh  th  of  40  ?     Of  8  ?     Of  48  ?    Of  72  ? 

8.  What  is  a  ninth  of  27  ?     Of  72  ?     Of  18  ?    Of  36  ? 

9.  If  7  pounds  of  rice  cost  56  cents,  what  does  1  pound 
cost? 

10.  If  9  tons  of  ice  cost  $63,  what  does  1  ton  cost  ? 

11.  If  9  oranges  cost  36  cents,  what  does  1  orange  cost? 

12.  A  lady  gave  16  apples  to  4  boys,  giving  to  eaxih  boy 
the  same  number.    How  many  apples  did  each  boy  receive  ? 

13.  If  4  quarts  of  milk  cost  24  cents,  how  much  does 
1  quart  cost  ? 

14.  If  8  pencils  cost  48  cents,  what  does  1  pencil  cost? 

15.  If  8  bushels  of  coal  cost  64  cents,  what  does  1  bushel 
cost? 

16.  A  boy  divides  84  cherries  equally  among  7  boys.  How 
many  does  he  give  to  each  ? 


66 


INTEGERS. 


t^Wi^itten  ^:?^erei<esr42 


CASE    I. 
105.  When  the  Divisor  does  not  exceed  9, 


Example.— Divide  6716  by  4. 

SOLUTION. 

Divisor,   4)6716   Dividend. 


Explanation. — We  write 
the  divisor  at  the  left  of  the 
dividend,  with  a  curved  line 
1679    Quotient.  between  them. 

As  the  first  figure  of  the 
dividend,  6,  is  thousands,  we  divide  6  by  4,  which  gives  us  1  thousand, 
and  2  thousands  remainder  (70). 

3  thousands  equal  20  hundreds  ;  to  which  we  add  the  7  hundreds  of 
the  dividend  and  obtain  27  hundreds. 

Dividing  27  hundreds  by  4,  we  have  6  hundreds,  and  3  hundreds  or 
30  tens  remainder.  Adding  the  30  tens  to  the  1  ten  of  the  dividend, 
we  have  31  tens. 

Dividing  31  tens  by  4,  we  have  7  tens,  and  3  tens  or  30  ones 
remainder.  Adding  the  30  ones  to  the  6  ones  of  the  dividend,  we 
have  36  ones. 

Dividing  36  ones  by  4,  we  have  9  ones. 

We  have  now  used  all  the  figures  of  the  dividend,  and  the  result, 
1679,  is  the  quotient  required, 

106.  Rule.—  Write  the  divisor  at  the  left  of  the  dividend, 
separating  them  hy  a  curved  line. 

Find  hoiu  many  times  the  divisor  is  contained  in  the  left- 
hand  figure  or  figures  of  the  dividend,  and  write  the  result  as 
apart  of  the  quotient.  Consider  the  remainder,  if  any,  as 
'prefixed  to  the  next  figure  of  the  dividend  ;  divide  as  before, 
and  write  the  result  as  the  second  figure  of  the  quotient. 
Proceed  thus  with  all  the  figures  of  the  dividend. 


DIVISION^.  67 

Should  any  figure  of  the  dividend,  except  the  first,  not  con- 
tain the  divisor,  ivrite  a  cipher  in  the  same  order  vn  the  quo- 
tient and  proceed  loith  the  figure  as  a  remainder, 

Pkoof. — The  'product  of  the  quotient  and  divisor  should 
equal  the  dividend. 

PROBLEMS, 

1.  How  many  times  6  in  486018  ?  Ans,  81003. 

2.  How  much  is  one-sixth  of  1598524  ?      Ans.  $99754. 

3.  How  many  times  7  in  567014  ?  Ans.  81002. 

4.  How  much  is  one-seventh  of  14813571. 

Ans.  $687653. 

5.  How  many  times  8  in  7901144?  Ans.  987643. 

6.  How  much  is  one-eighth  of  $803448  ?  Ans.  $100431. 

7.  How  many  times  9  in  630927  ?  Ans.  70103. 

8.  How  much  is  one-ninth  of  $282609  ?      Ans.  $31401. 

9.  If  apples  are  $2  a  barrel,  how  many  barrels  can  be 
bought  for  $1970  ?  Ans.  985  barrels. 

10.  If  6  acres  of  land  cost  $1656,  what  is  the  cost  of  one 
acre  ?  Ans.  $276. 

11.  If  flour  cost  $8  a  barrel,  how  many  barrels  can  be 
bought  for  $73264?  Ans.  9158  barrels. 

12.  Divide  $5872  equally  among  4  men.       Ans.  $1468. 

13.  There  are  7  days  in  one  week.  How  many  weeks  are 
there  in  36512  days  ?  Ans.  5216  weeks. 

14.  Three  feet  make  one  yard.  How  many  yards  are 
there  in  4701045  feet  ?  Ans.  1567015  yards. 

15.  Divide  an  estate  of  $93760  equally  among  5  heirs. 

Ans.  $187^2. 

16.  At  9  miles  an  hour,  in  how  many  hours  would  a  ship 
sail  2925  miles  ?  Ans.  325  hours. 


68 


INTEGERS 


17.  A  man  whose  wages  were  $4  a  day,  earned  $1148. 
How  many  days  did  he  work  ?  A  ns.  287  days. 

18.  If  3  horses  eat  2115  pounds  of  hay  in  a  month,  how 
much  will  one  horse  eat  ?  Ans.  705  pounds. 

19.  A  builder  paid  $2580  for  bricks,  at  $6  a  thousand. 
How  many  thousand  did  he  buy?        Ans.  430  thousand. 

20.  A  party  of  9  men  spent  $7245  on  a  journey  to  Europe, 
sharing  the  expenses  equally.  How  much  did  each  man 
pay?  Ans.  $805. 

21.  How  long  will  it  take  a  boat  to  make  a  voyage  of 
749  miles,  if  she  travels  at  the  rate  of  7  miles  an  hour  ? 

Ans.  107  hours. 

22.  A  man  leased  a  farm  for  $1500,  at  the  rate  of  $6  an 
acre.    How  many  acres  in  the  farm  ?         Atis.  250  acres. 


^-IWritten^TO^rci^eX* 


CASE    II. 
107.  When  the  Divisor  is  greater  than  9. 


Example.— Divide  37245  by  65. 


Divisor.  Diridend. 

Quotient. 

65 

)  37245  (  573 

325 

474 

Partial  Dividend. 

455 

195 

Partial  Dividend. 

195 

Explanation. — We  write  the 
dividend  and  divisor  as  in  Case  I. 

The  smallest  number  ex- 
pressed by  the  left  hand  figures 
of  the  dividend,  and  which  will 
contain  the  divisor  is  372  hun- 
dreds, which  we  divide  by  65, 
obtaining  5  hundreds  for  the 
first  figure  of  the  quotient.  We 
multiply  65  by  5  hundreds  and 


DIVISION". 


69 


obtain  325  hundreds,  which  subtracted  from  373  hundreds,  leave  47 
hundreds  =  470  tens.  To  this  we  add  the  4  tens  of  the  dividend,  or 
which  is  the  same  thing,  annex  the  4  tens  to  the  47  hundreds,  and  we 
have  474  tens  for  a  new  dividend,  ov partial  dividend,  because  it  is. a 
part  of  the  original  dividend. 

65  is  contained  in  474  tens  7  tens  times,  and  we  write  the  7  tens  in 
the  quotient.  65  multiplied  by  7  tens  =  455  tens ;  and  474  tens  — 
455  tens  leaves  19  tens  —  190  ones.  Adding  the  5  ones  of  the  divi- 
dend, we  have  for  a  new  dividend  195  ones. 

65  is  contained  in  195  ones  3  times.  We  write  3  in  the  ones'  place 
in  the  quotient.  65  x  3  =  195.  195  from  195  leaves  no  remainder,  and 
the  required  quotient  is  573. 

Example.— Divide  54397  by  132. 

SOLUTIONS. 


First  Method. 

Second  Method. 

Divisor.  Dividend.  Quotient. 

Dividend. 

132  )  54397  ( 

412 

54397  (  132 

Divisor. 

528 

528      (  412 

Quotient. 

159 

159 

132 

132 

277 

277 

264 

264 

13 

Remainder. 

13      Remainder. 

PROOF. 

412 
-  132 

Quotient. 
Divisor. 

>  CaUed  Factors  in  MultipUcation. 

824 

1236 

412 

54384 
13 

54397 

108.  Rl'LE. —  Write  the  divisor  at  the  left  or  right  of  the 
dividend^  luith  a  curved  line  hettoeen  them. 


Remainder. 

Dividend.    Called  Product  in  Multiplication. 


70 


INTEGERS. 


Separate  mentally  from  the  left  of  the  dividend  the  figures 
expressing  the  smallest  number  that  will  contain  the  divisor. 
Divide  this  nuniber  hy  the  divisor,  and  write  the  result  as  the 
first  figure  of  the  quotient. 

Multijjly  the  divisor  hy  this  quotient  figure,  and  subtract 
the  product  from  that  part  of  the  dividend  which  was  used. 

To  the  right  of  the  reynainder,  if  there  is  any,  annex  the 
r.ext  figure  of  the  dividend,  and  divide  the  mimber  thus 
formed  by  the  divisor,  ivriting  the  result  as  the  second  figure 
of  the  quotient. 

'Proceed  in  this  manner  until  all  the  figures  of  the  dividend 
have  been  used. 

If  any  remainder  with  the  next  figure  of  the  dividend 
annexed,  does  not  contain  the  divisor,  write  a  cipher  in  the 
quotient,  annex  to  the  remainder  another  figure  of  the  divi- 
dend, and  proceed  as  before. 

Proof. — Multiply  the  divisor  and  quotient  together,  and 
to  the  product  add  the  remainder,  if  any.  If  the  result  is 
equal  to  the  dividend,  the  ivorh  is  correct 


P  ROBI.EM, 


1.  67473 

2.  22491 

3.  44982 

4.  179928 

5.  112455 

6.  157437 

7.  202419 

8.  292383 

9.  314874 
10.  472311 


a  13 

b  14 

c  15 

d  16 

e  18 

/  ^1 

9  24 

h  27 

i  32 

j  34 


Explanation.— Form  100 
Problems  by  dividing  each 
of  the  dividends  1,  2,  3,  4,  5, 
etc.,  by  each  of  the  divisors 
a,  b,  c,  d,  e,  etc.,  as 

\a.  67473-7-13  =  ? 

\g.  67473^24  =  ? 

6a.  157437-f-13  =  ? 

6d.  157437-^16  =  ? 


DIVISIOK. 


71 


11.  Divide  three  million  twenty-five  thousand  six,   by 
four  thousand  eighteen.  Remainder,  3470. 

12.  In  638  days  a  government's  revenue  was  $246356682. 
How  much  was  that  per  day  ?  Ans.  1386139. 

13.  If  a  railroad  earns  $2600000  in  52  weeks,  how  much 
is  that  per  week  ?  Ans.  $50000. 

14.  A  railroad  357  miles  long  cost  $2502927.    How  much 
was  that  per  mile  ?  Ans.  $7011. 

15.  In  one  gallon  of  wine  there  are  231  cubic  inches. 
How  many  gallons  in  8547  cubic  inches  ?   Ans.  37  gallons. 

16.  Sound  travels   1119  feet  in  1   second.     How  long 
would  it  take  it  to  travel  one  mile,  or  5280  feet  ? 

Ans.  4  seconds,  and  804  feet  remainder. 


■^ 


^Written  ^xfei'ci^e*^^ 


CASE    III 

109.   WJien   the   Divisor   is 
annexed. 


1   with   ciphers 


Example.— Divide  1462  by  100. 

Explanation. — We  divide  accord- 
ing to  Explanation  and  Rule  in 
107  and  108,  and  obtain  tlie  quo- 
tient 14,  and  remainder  63. 

We  can  obtain  tlie  same  result  by 
a  much  shorter  process,  name'y, 
cutting  off  the  two  right-hand  fig- 
ures of  the  dividend,  thus  14/62,  in 
wliich  we  have  14  for  the  quotient, 
and  62  for  the  remainder  as  before. 


solution. 
Divisor.  Dividend.  Quotient. 

100  )  1462  (  14 
100 
462 
400 

62  Remainder. 


Example.— Divide  17563  by  1000. 
Solution.— 17563-5-1000  =  17,  with  563  remainder. 


73  INTEGERS. 

Explanation. — We  cut  off  from  tlie  right,  tliree  figures  (the  num- 
ber of  ciphers  in  the  divisor)  and  have  14  as  the  quotient,  and  563.  the 
remainder. 

In  any  number  all  the  figures  to  the  left  of  ones,  express  tens ; 
to  the  left  of  tens,  hundreds ;  to  the  left  of  hundreds,  thousands ; 
etc.  Therefore,  in  the  number  17563,  17  must  express  17  thousands, 
and  563  must  be  the  remainder. 

110.  EuLE. — Cut  off  from  the  right  of  tlie  dividend  as 
many  figures  as  there  are  ciphers  at  the  right  of  the  divisor. 
The  remaining  figures  form  the  quotient,  and  the  figures  cut 
off,  the  remainder. 

PROBLEMS. 

1.  Divide  48729  by  100 ;  by  1000 ;  by  10000. 

2.  Divide  637591  by  1000 ;  by  100000 ;  by  10. 

3.  Divide  1234567  by  1000000 ;  by  10000. 

111.  When  the  divisor  is  any  numler  with  ciphers 
annexed,  the  following  Kule  answers^ our  purpose  : 

KuLE. —  Cut  off  from  the  right  of  the  dividend  as  many 
figures  as  there  are  ciphers  at  the  right  of  the  divisor. 
Then  divide  the  remaining  figures  in  the  dividend  hy  the 
divisor  exclusive  of  its  right-hand  ciphers. 

To  the  remainder,  if  any  after  this  division,  annex  the 
figures  cut  off  from  the  dividend  for  a  true  remainder. 


Example.— Divide  2400 
by  60. 

SOLUTION. 

6|0 )  240|0 
40 

Example.— Divide   13762 
by  400. 

SOLUTION. 

4!00)137|62 

34  quotient,  162  remainder. 

PBOB 

Ex.  1.  Divide  3472  by  40. 

2.  Divide  476375  by  70. 

3.  Divide  167478  by  90. 

ZEMS. 

4.  Divide  17560  by  400. 

5.  Divide  715723  by  700. 

6.  Divide  752562  by  340. 

SECTION    VII. 
,^» 


MWi'itten  ^^erci^es^f 


MEVIEW    PHOBIjEMS. 

1.  The  addends  of  a  number  are  64,  92,  430,  7859,  and 
436177.     What  is  the  number  ?  Ans.  444628. 

2.  The  parts  of  a  number  are  4876,  73,  23756,  400009, 
and  99999.     What  is  the  number?  Ans.  528713. 

3.  The  minuend  is  48639  and  the  subtrahend  is  29048. 
What  is  their  difference?  Ans.  19591. 

4.  The  minuend  is  93470  and  the  subtrahend  is  39401. 
What  is  the  remainder  ?  Ans.  54069. 

5.  The  minuend  is  47319  and  the  difference  is  12499. 
What  is  the  subtrahend  ?  Ans.  34820. 

6.  The  minuend  is  311409  and  the  remainder  is  76234. 
What  is  the  subtrahend  ?  Ans.  235175. 

7.  The  subtrahend  is  4008  and  the  difference  is  17296. 
What  is  the  minuend  ?  Ans.  21304. 

8.  The  subtrahend  is  93000  and  the  remainder  is  639421. 
What  is  the  minuend?  Ans.  732421. 

9.  The  multiplicand   is    4215   and  the  multiplier  309. 
What  is  the  product?  Ans.  1302435. 

10.  The  multiplicand  is  937  and  the  product  is  40291. 
What  is  the  multiplier?  Aiis.  43. 

11.  The  multipher  is  319  and  the  product  is  320276. 
What  is  the  multiplicand  ?  Ans.  1004. 


74  INTEGERS. 

12.  The  dividend  is  382268  and  the  divisor  is  421.     What 
is  the  quotient  ?  Ans.  908. 

13.  The  dividend  is  387234  and  the  quotient  is  909. 
What  is  the  divisor?  Ans.  426. 

14.  The  divisor  is  145  and  the  quotient  is  4200.    What 
is  the  dividend  ?  Ans.  609000. 

15.  The  divisor  is  43,  the  quotient,  64,  and  the  remainder, 
15.    What  is  the  dividend  ?  Ans.  2767. 

16.  The  dividend  is  2943  and  the  divisor  is  66.    What  is 
the  remainder  ?  Ans.  39. 

17.  The  dividend  is  9600  and  the  quotient  is  95.     What 
is  the  remainder?  Ans.  5. 

18.  From  the  sum  of  72  and  83  take  the  difference  of 
125  and  182.  Ans.  98. 

19.  (127  +  468)  —  (1930  —  1678)  =  ?         Ans.  343. 

20.  To  the  product  of  45  by  23  add  the  quotient  of  512 
by  16.  Ans.  1067. 

21.  ( 93  X  32 )  4-  ( 1462  -^  43 )  =  ?  Ans.  3010. 

22.  Multiply  the  sum  of  83  and  74  by  their  difference, 
and  divide  the  product  by  15.  Ans.  94,  3  remainder. 

23.  (99  +  48)  X  (99  _  48)  -  7  H-  35  =  ? 

Ans.  214. 

24.  Multiply  the  quotient  of  3348  divided  by  27  by  the 
quotient  of  150000  divided  by  375.  A?is.  49600. 

25.  (1287  -T-  99)  X  (30969  -^  999 )  =  ?      Ans.  403. 

26.  Bought  160  acres  of  land  at  $5  an  acre  and  sold  it  at 
$12  an  acre.    What  did  I  gain  ?  Ans.  $1120. 

27.  140  X  (16-7)=?     (140  x  16)-  (140  x  7)=? 

Ans.  1260. 

28.  Bought  375  barrels  flour  for  13000;  sold  them  for 
$2260.    How  much  per  barrel  did  I  lose  ?  Ans,  $2. 


REVIEW     PROBLEMS.  75 

29.  (3600  -  1350) -^  450  =  ?  ( 3600 -^  450)  -  (1350 — 
450)  =  ?  Ans.  5. 

30.  Bought  5  yards  of  cloth  at  $7  a  yard,  3  yards  of  silk 
at  $2  a  yard,  and  12  shirts  at  $3  apiece,  and  offered  four  $20 
bills.     What  change  was  due  me  ?  Ans.  $3. 

31.  176  -  (9x7)  +  (5  X  8)  +  (14  x  5)  =  ? 

Ans.  3. 

32.  B's  income  one  year  was  $2500;  his  expenses  that 
year  averaged  $3  per  day :  how  much  did  he  save,  the  year 
having  365  days.  Ans.  $1405. 

33.  10000  —  (421  X  13)  =  ?  Ans.  4527. 

34.  A  farmer  had  $5,  and  sold  his  marketing  for  $28 ;  he 
bought  $4  worth  of  sugar,  $3  worth  of  coffee,  and  $2  worth 
of  spices.    How  much  had  he  left?  Ans.  $24. 

35.  (1900  +  396)  —  (561  +  411  +  96)  =  ? 

Ans.  1228. 

36.  Mr.  Hall,  keeping  his  store  open  310  days  in  1  year, 
made  $3100;  and  his  family  expenses  for  365  days  were 
11095.  How  much  did  he  make  per  working  day  more 
than  he  spent  per  spending  day  ?  Ans.  $7. 

37.  (20736  ^  144)  —  (1728  -^  144)  =  ?     Ans.  132. 

38.  How  many  times  can  a  40-gallon  barrel  be  filled  from 
5  casks,  each  holding  120  gallons  ?  Ans.  15  times. 

39.  1700  X  400  -^  34000  =  ?  Ans.  20. 

40.  Bought  300  barrels  of  apples  at  $3  per  barrel ;  120 
of  them  were  stolen :  at  what  price  per  barrel  must  sell  the 
remainder  so  as  to  suffer  no  loss?  Ans.  $5. 

41.  (1900  X  20)  -f-  (1900  -  1140)  =  ?       Ans.  50. 

42.  Five  thousand  six  hundred  seventy-eight,  and  another 
number,  amount  to  twenty-five  thousand  four.  What  is  the 
other  number?  Ans.  19326. 


76 


INTEGERS. 


43.  A  public  school  pays  its  principal  11800  per  annum; 
2  assistant  principals  1800  each;  6  medium  teachers  $600 
each ;  and  8  primary  teachers  $500  each.  What  is  the  cost 
of  instruction  ?  Ans.  $11000. 

44.  2500  +  (3  X  500)  +  (5  x  700)  +  (6  X  200)  =? 

Ans.  8700. 

45.  If  a  mechanic  receives  $1400  a  year  for  his  labor,  and 
his  expenses  are  $840,  in  what  time  can  he  save  enough  to 
buy  32  acres  of  land  at  $140  an  acre  ?  Ans.  8  years. 

46.  (280  X  64)  -f-  (2504  —  1384)  =  ?  Ans.  16. 

47.  A's  age  is  35,  and  B's  27  years.     What  is  the  average 

of  their  ages  ? 

Operation. -7- (35  years  +  27  years)  -4-  2  =  81  years. 
The  average  of  two  numbers  is  one-half  their  sum  ;  the  average  of 
three  numbers,  one-third  their  sura,  &c. 

48.  Find  the  average  of  2,  4,  and  6.  Ans.  4. 

49.  Find  the  average  of  3,  5,  and  10.  Ans.  6. 

50.  Find  the  average  of  6,  10,  and  11.  Ans.   9. 

51.  Five  men  are  worth,  respectively,  $3000,  $12000, 
$56000,  $20000,  and  $35000.  What  are  they  worth  on  the 
average  ?  Ans.  $25200. 

52.  A  boy's  earnings  per  month  were  as  follows:  $12,  $10, 
$15,  $13,  $16,  $18,  $14,  $8,  $20,  $17,  $15,  and  $10.  What 
were  his  average  earnings  per  month  that  year?    Ans.  $14. 

53.  Farmer  A's  expenses  and  receipts  one  year  were  as 
follows : 


EXPENSES, 

RECEIPTS. 

^01 

'  Labor     .... 

$315 

For  Wheat    .    .     . 

.  $364 

(( 

Seed 

64 

"    Corn  .... 

.     140 

it 

Stock     .     .     .    . 

290 

"    Pork  .... 

.     216 

(( 

Fruit  Trees    .     . 

100 

"    Fruit.    .     .     . 

.      79 

u 

Family  Expenses 

590 

"Hay   ...    . 

.     195 

a 

A  Reaper    .     .    . 

375 

«    Vegetables.     . 

.       46 

REVIEW      PROBLEMS. 


Did  he  make  or  lose  that  year,  and  how  much  ? 

Ans.  Lost  $694. 
54.  Farmer  B's  expenses  and  receipts  one  year  were  as 
follows : 


EXPENSES. 

For  Labor    ....   $516 

"    Seed 309 

"    Stock     ....       50 
"    A  Mower   ...     190 
"    Family  Expenses      478 
Did  he  make  or  lose  that  year,  and  how  much  ? 

Ans.  Made  $194. 
55.  Farmer  C's  expenses  and  receipts  one  year  were  as 
follows : 

RECEIPTS.  EXPENDITURES. 


RECEIPTS. 

For  Grain  .  . 
"  Wool  .  . 
"  Beef  .  .  . 
«    Hay  .     .    . 

"    Vegetables 


$816 

400 

180 

295 

46 


For  Grain  . 
"    Stock  . 


.     .  $4005         For  Seed $145 

.    .       984  "    Labor     ....     875 

Did  he  gain  or  lose  money,  and  how  much  ? 

Ans.  Gained  $3969. 
56.  The  sum  of  two  numbers  is  72,  and  their  difference 
is  46.     What  are  the  numbers  ? 


EXPLANATION. 

Since  72  is  the  sum  of  the  numbers, 
if  the  difference  46  is  subtracted  from 
the  sum  72,  the  remainder  26  will  be 
twice  the  less  number. 


SOLUTION. 

72- 

46 

=  26. 

26h- 

2 

=  13,  Less  No. 

13  + 

46 

=  59,  Greater  No. 
PROOF. 

59  -f 

13 

=  72,  Sum. 

59- 

13 

=  46,  Difference. 

72  +  46 

=  118. 

118 -^ 

2 

=     59,  Greater  No. 

59- 

46 

=     13,  Less  No. 

26  -i-  2  =  13,  the  less  number,  and 
13  +  46  =  59,  the  greater  number. 

Or,  if  the  difference  46  is  added  to 
the  sum  72,  the  amount  118  will  be 
twice  the  greater  number. 

118  ^  2  =  59,  the  greater  number, 
and  59  —  46  =  13,  the  less  number. 


78  INTEGERS. 

57.  The  sum  of  two  numbers  is  1196,  and  their  difference 
is  504.     What  are  the  numbers  ? 

A71S.  Greater,  850;  Less,  346. 

58.  The  sum  of  two  numbers  is  8356,  and  their  difference 
is  1792.     What  are  the  numbers  ? 

Ans.  Greater,  5074;  Less,  3282. 

59.  A  man  paid  14200  for  a  house  and  lot,  the  lot  being 
valued  at  $1950  more  than  the  house.  What  was  the  value 
of  each?  Ans.  Lot,  $3075;  House,  11125. 

60.  Two  men  are  worth  $43875,  and  one  is  worth  $2965 
more  than  the  other.    What  is  each  man  worth  ? 

A71S.  1st  man,  $23420;  2d,  $20455. 

The  pupils  will  make  original  examples  to  illustrate  the  following 
problems : 

61.  Given  the  sum  and  all  the  parts  but  one,  to  fin  1  that 
one. 

62.  Given  the  greater  of  two  numbers  and  their  difference, 
to  find  the  less. 

63.  Given  the  product  and  two  of  three  factors,  to  find  the 
otJier  factor. 

64.  Given  the  product  and  multiplier,  to  find  the  multi- 
plicand. 

65.  Given  the  divisor,  quotient,  and  remainder,  to  find 
the  dividend. 

66.  Given  the  sum  and  difference  of  two  numbers,  to  find 
the  numbers. 

67.  Given  the  divisor  and  quotient,  to  find  the  dividend. 

68.  Given  the  product  and  one  of  two  factors,  to  find  the 
other  factor. 

69.  The  pupils  will  make  two  original  examples  in  each 
of  the  fundamental  rules. 


OUTLINE    OF    DECIMALS. 


125 
137 

139 


1st  METHOD.    < 


113.  Decimal  Point. 

114.  Value. 


PLACES. 


113.  Point. 

115.  Tenths. 

116.  Ilundvedths. 

117.  Tlionsandths. 
&c.,  &c.t  &c. 


118  and  119.  Bnles. 


2d  METHOD, 


ntiyciPLES. 


(    120. 
X  121 

(   133. 
I    124. 


Decimal  Point, 
and  122.  Rules. 

First. 
Second. 


.  ADDITION  ...VZ6.  Ttule. 

.  .Sl7JJTiJ4CTJC»iV....128.  JJuZe. 


.  MTTLTIPLI- 
CATION. 


DIVISION, 


PRINCIPLES 
132.  JRule. 


PRIKCIPLES. 


■1 


130.  1st. 

131.  2d. 

133.  1st. 

134.  2d. 

135.  Sd. 

136.  4</i. 

137.  5th. 


139.  17.  «.  MONEY,  < 


138.  iSule. 

140.  Table. 

141.  I7n«. 

142.  ^  Numeration 

143.  V  and 

144.  )      Notation. 

145.  ^ddJition. 

146.  Subtraction. 

147.  Multiplication. 

148.  l>it'i«ion.. 

7» 


CHAPTER    II.. 


~§fr(^ 


J0)EJCT[MA1L,§; 


II 


(jg-^v^ 


SECTION    I 


'^-^ 


IMBMEIRJ^IIOM  smm  N©TA11®M 


(i^'^ 


«^s^ 


From  the   Outline  of  Arithmetic,  pag^e  11,  we  learn  that 
numbers  are  classified  as  to  wholeness  of  their  unit  into  Integral,  Frac- 
tional, and  Mixed  Numbers. 
8C 


NUMEEATION     AKT>     NOTATION.  81 

We  have  already  considered  Numeration  and  Notation,  Addition^ 
Subtraction,  Multiplication,  and  Division  of  Integers,  and  are  now 
prepared  to  proceed  with  Fractions. 

A  Fraction  ( 7 )  is  a  number  expressing  one  or  more  of  the 
equal  parts  of  a  unit ;  dL^five  tenths,  one  half. 

Fractions  with  reference  to  the  mode  of  expressing  them  are  of  two 
kinds,  namely.  Decimal  and  Common. 

112.  Decimals  are  those  fractions  which  express  one 
or  more  of  the  parts  of  a  unit  which  has  been  divided  into 
tenths,  hundredths,  thousandths,  &c. 

Decimals  are  expressed  in  accordance  with  the  Principles  of  Arabic 
Numeration  and  Notation  of  Integers  (41). 

The  term  decimal  is  derived  from  the  Latin  decern,  which  signifies  10. 

A  number  consisting  of  an  Integer  and  Decimal  is  a  Mixed  Num- 
lev  (8). 

By  referring  to  the  picture  on  page  80,  you  will  notice  a  ball  resting 
on  the  keystone  (units)  of  the  arch.  This  represents  the  Decimal 
Point,  and  is  in  decimals  yfhaX  ones  of  units  are  in  inlegerSi—ihet  First 
Place. 

113.  The  Decimal  Point  is  used  to  distinguish 
Decimals  from  Integers. 

When  the  point  is  placed  between  figures  it  is  read  and  ;  thus,  2.5  is 
read  "  2  and  5  tenths." 

It  may  also  be  seen  in  the  picture  that  places  equally  distant  from 
units  have  corresponding  names,  with  the  exception  of  the  decimal 
termination  ths.     Thus,  tens, — tenths  \    hundreds, — hundred^A*;    &c 


114.   The  Value  which  any  figure  in  a  decimal  ex- 
presses is  determined  by  the  'place  it  occupies. 

The  Places  are  Point,  TentTis,  Hundredths,  ThonsandtJis,  Ten-thov^ 
sandths,  &c. 

6 


5  DECIMALS. 

115,  Tenths  is  the  Second  Place  or  Order. 

.1  rzr  1  tenth.  .4  =  4  tenths.         .7  =  7  tenths. 

.2  =  2  tenths.  .5  =  5  tenths.  .8  =  8  tenths. 
.3  =  3  tenths.  .Q  =  Q  tenths.  .9  =  9  tenths. 
The  greatest  number  of  tenths  expressed  ly  one  figure  is  9. 

10  tenths  =  1. 


EXERCISES. 


1.  Read  .6;  6.6;  .9;  9.9;  18.7;  23.5. 

2.  Write  eight  tenths;  12  and 3  tenths;  745  and  4  tentha 

116.   hundredths  is  the  Third  Place  or  Order. 


.01  =  1  hundredth. 
.02  =  2  hundredths. 
.03  =  3  hundredths. 
.04  =  4  hundredths. 


.11 
.32 
.53 

.84 


11  hundredths. 
32  hundredths. 
53  hundredths. 
84  hundredths. 


The  greatest  number  of  hundredths  expressed  by  two  figures  is  99. 

100  hundredths  =  1. 


EXERCISES. 

1.  Read  .24;  5.47;  .88;  475.15. 

2.  Read  42.09;  .75;  1876.93;  .19. 

3.  Read  14.14;  23.23;  44.44;  .07. 

4.  Read  111.11;  213.02;  400.04;  500.05. 

5.  Write  one  hundred  and  five  hundredths. 

6.  Write  fifty  and  forty-one  hundredths. 

7.  Write  seventy-six  hundredths;  10  hundredths. 

8.  Write  nine  hundred  ninety-nine  and  ninety-nine  hun- 
dredths. 

9.  Write  three  thousand  forty  and  forty  thousand  three 
h  undred-thousandths. 


KUMERATIOX     AND     NOTATIOiq-.  83 

117.  Thousattdths  is  the  Fourth  Place  or  Order. 
.001  =  1  thousandth.  I       .098  =  98  thousandths. 
.014  =  14  thousandths.  .942  =  942  thousandths. 
.144  =  144  thousandths.      |       .004  =  4  thousandths. 
2%e  greatest  number  of  thousandths  expressed  by  three  figures  is  999. 

1000  thousandths  =  1. 

±:XEliCISES. 

1.  Read  .359 ;  4000.004 ;  35.035. 

2.  Eead  485.485 ;  75.075 ;  19000.019. 

3.  Read  .011;  408.008;  143.043. 

4.  Read  4279.279;  11.048;  .099. 

5.  Write  325  thousandths;  4  thousandths. 

6.  Write  97  and  12  thousandths;  803  thousandths. 

7.  Write  4  and  4  thousandths;  47  thousandths. 

8.  Write  11245  and  45  thousandths;  9  thousandths. 

When  the  right-hand  figure  of  a  decimal  is  handredths  ( 1 16),  the 
decimal  is  read  as  hundredths;  when  the  right-hand  figure  is  thou- 
sandths (117)  the  decimal  is  read  as  thousandths ;  &c.     Hence, 

118.  Rule  for  Numeratoi?'.  —  Read  the  figures  of  a 
decimal  the  same  as  the  figures  of  an  integer ,  adding  the 
name  of  the  right-hand  order  of  the  decimal. 

119.  Rule  for  Notation. —  Write  the  given  decimal  as 
a  whole  number.  Prefix  ciphers,  if  necessary,  to  fill  the  re- 
quired number  of  places.    To  the  left  write  the  decimal  point. 

Example.  — Read  123456.72345. 

By  the  preceding  method  we  are  required  to  perform  three  operar 
tions  in  order  to  read  the  example  given : 

First.— To  read  the  integer— if,?  thousand  J^56. 

Second. — To  read  the  decimal  as  an  integer — and  72  thousand 
345. 

Third. — To  ascertain  the  name  of  the  right-hand  decimal  place  by 
beginning  at  7  and  saying  tenths,  hundredths,  thousandths,  ten-thou- 
sandths, hundred-thousandths. 


84 


DECIMALS. 


SECOJfD    METHOD. 

120.  The  Decimal  I^oint  ( . )  occupies  that  place  in 
a  decimal  number  which  gives  narm  to  the  decimal. 

Example.  — Read  123456.72345. 

Integer.      •      Decimal. 

123  45?i^34? 

Thous.  Units. 

I23  45?  =  123  thousand  4S6  and 

Thous.  Cnits. 

.72  345  =  72  thousand  343  hundred-thou^ 
sandths. 

By  the  Second  method  the  integer  is  read  as  in  the  First  method. 
But  in  reading  the  decimal  as  an  integer  we  notice  that  the  point  is  in 
hundreds'  order  of  thousands'  period  and  know  instantly  that  hundred- 
thousandths  is  the  name  of  the  decimal.     Hence, 

121.  Rule  for  Numeration.  —  Read  the  figures  of  a 
decimal  the  sa7ne  as  the  figures  of  an  integer,  adding  the 
name  of  the  "place  in  the  decimal  occupied  hy  the  pointy  count- 
ing from  the  right  as  in  integers, 

122.  Rule  for  Notation.  —  Write  the  given  decimal 
as  a  whole  number,  prefixing  ciphers,  if  necessary,  to  cause 
the  point  to  stand  in  the  place  luhich  gives  the  name  of  the 
decimal. 

EXERCISES. 

Read  the  following: 


1.  .125  623 

2.  .400  119 

3.  .50  091 

4.  .62  124 

5.  .142  205 


6.  .423  152  391 

7.  .72  110  111 

8.  .00  095  486 

9.  .000  004  004 
10.    .400  092 


11. 
12. 
13. 
14. 


.125 

.0  125 

.00  125 

.4  125 


15.  .111  222 


NUMEEATIOi^  AND  IS^OTATION.        { 

16. 

421  234  .23  191 

27. 

1  004   .1  004 

17. 

75  629  .119  256 

28. 

93  093  .93  093 

18. 

40  000   .0  004 

29. 

175  175  .175  175 

19. 

914  .173  216  014 

30. 

42  420  .42  420 

20. 

400  000  .00  004 

31. 

750  000  .750 

21. 

50  000  .050 

32. 

400  000  .400 

22. 

10  000   .0  001 

33. 

27  516  .516 

23. 

19  019  .019 

34. 

44  550  .00  009 

24. 

128.000  000  128 

35. 

833  338  .000  009 

25. 

4.000  000  004 

36. 

.00  000  000  009 

26. 

934  761  132   .4 

37. 

.9  009  000  909 

85 


Write  the  following  in  figures: 

1.  Two  hundred  fifty-six  ten-thousandths. 

2.  Nineteen  thousand  one  millionths. 

3.  Fourteen  hundred-thousandths. 

4.  Eight  thousand  ninety-nine  billionths. 

5.  One  million  three  ten-millionths. 

6.  Seven  hundred  forty  thousand  one  millionths. 

7.  Ninety-eight  and  one  hundred  three  thousand  four 
ten-millionths. 

8.  Four  thousand  and  four  hundred-thousandths. 

9.  Four  thousand  four  hundred-thousandths. 

10.  Four  thousand  and  four  hundred  thousandths. 

11.  Six  thousand  and  six  hundred-thousandths. 

12.  Six  thousand  six  hundred-thousandths. 

13.  Six  thousand  and  six  hundred  thousandths. 

14.  Three  hundred  one  billionths. 

15.  Ninety-three  million  and  93  millionths. 

16.  Eleven  billion  and  eleven  hundred-millionths. 

17.  Five  tenths;  Fifty  hundredths;  Five  hundred  thou- 
sandths; Five  thousand  ten-thousandths. 


86  DECIMALS. 

18.  Seventy-five  hundredths. 

19.  Twenty-eight  thousandths. 

20.  One  hundred  nine  ten-thousandths. 

21.  Fifteen  and  thirty-two  hundred-thousandths. 

22.  Seven  and  three  hundred  five  milUonths. 

23.  The  decimal  two  thousand  six  ten-milhonths. 

24.  Two  thousand  six  ten-milUonths. 

25.  Six  thousand  nine  hundred- milUonths. 

26.  Six  thousand  9  hundred-millionths. 

27.  Three  hundred  one  billionths. 

28.  Three  hundred  and  one  billionth. 

29.  Two  hundred  and  five  ten-thousandths. 

30.  Forty  thousand  and  thirty-four  millionths. 

31.  Two  thousand  and  four  hundred-thousandths. 

32.  Six  hundred  and  fifteen  ten-millionths. 

33.  Six  hundred  units  and  fifteen  ten -thousandths. 

34.  Fifteen  and  fifteen  thousandths. 

35.  Three  hundred  thousand  three  hundred  and  three 
hundred-millionths. 

36.  Five  million  and  eighty-five  ten-millionths. 

37.  Twelve  hundred-thousandths. 

38.  Four  hundred  units  and  four  hundred  sixty-five 
millionths. 

.5  =  5  tenths.  .50  =  50  hundredths,  or  5  tenths  and  0  hundredths. 
.500  =  500  thousandths,  or  5  tenths  0  hundredths  and  0  thousandths. 
From  this  we  learn  that  .5,  .50,  and  .500  are  of  the  same  value— 5 
tenths.    Hence, 

1^3.  Ciphers  may  he  annexed  to  a  decimal  tvitJiout 
changing  its  value. 

124.  Ciphers  may  he  omitted  from  the  right  of  a  deci- 
mal without  changing  its  value. 


SECTION    II, 


«^|Mli&il^aiS1gMtet^ 


125.  Example.  — What  is  the  sum  of  43.249,  123.14, 

and  93.412? 

Explanation. — Since  only  like  orders  of  units  can  be 
added,  we  write  the  nambers  with  like  orders  in  the 
^*  same  columns. 

123.14  The  decimal  points  then  stand  in  a  column. 

93.412  Commence  at  the  right  and  add  as  in  integers,  plac- 

9^Q  «ni        ^°^  ^^^®  decimal  point  in  the  sum  directly  under  the 
decimal  points  in  the  parts. 

126.  KuLE. —  Write  the  numbers  so  that  the  decimal  points 
shall  stand  in  a  column.  Add  as  in  integers^  and  place  the 
point  in  the  sum  underneath  the  points  in  the  addends, 

PnOBIs  EMS. 

Find  the  sum 

1.  Of  6.09.5,  4.94,  8.573,  and  2.6.  Ans.  22.208. 

2.  Of  7.97,  .0568,  .04,  and  5.6378.  Ans.  13.7046. 

3.  Of  .853,  .0609,  9.7,  and  .00357.         Ans.  10.61747. 

4.  Of  .00328,  .6,  .88,  6.5,  and  .987654.    Ans.  8.970934. 

5.  Of  four  and  five  tenths,  forty-five  hundredths,  fifty- 
eight  thousandths,  eight  thousand  eight  ten-thousandths, 
seven  tenths,  and  ninety-six  millionths.     Ans.  6.508896. 

6.  Of  5  and  9  tenths,  4  and  58  hundredths;  6  and  29 
thousandths,  and  5708  millionths.  A^is.  16.514708. 


8S  DECIMALS. 

7.  On  the  Pa.  E.  K.  the  distance  from  Pittsburgh  to 
Wilkinsburg  is  6.9  miles;  from  Wilkinsburg  to  Greensburg 
is  24.2  miles ;  from  Greensburg  to  Johnstown,  46.9  miles ; 
from  Johnstown  to  Altoona,  38.7  miles;  from  Altoona  to 
Harrisburg,  129.3.  What  is  the  distance  from  Pittsburgh 
to  Harrisburg  ?  Ans.  246  miles. 

8.  On  the  P.  W.  &  B.  R  K.  the  distance  from  Pittsburgh 
to  Braddocks  is  9.6  miles;  from  Braddocks  to  West  Newton, 
23.1  miles;  from  West  Newton  to  Confluence,  51.5  miles; 
from  Confluence  to  Cumberland,  65.3  miles.  What  is  the 
distance  from  Pittsburgh  to  Cumberland  ? 

Ans.  149.5  miles. 

9.  A  farmer  in  buildiug  a  fence  around  his  farm  found 
that  the  sides  measured  respectively  28.17  rods,  30.485  rods, 
25.672  rods,  68.875  rods,  and  75.125  rods.  How  many  rods 
of  fence  did  he  have  to  build  ?  Ans.  228.327  rods. 

10.  A  dealer  in  Western  lands  sold  to  one  party  325.75 
acres ;  to  a  second,  475.375  acres ;  to  a  third,  37.565  acres ; 
and  to  a  fourth,  379.598  acres.  How  many  acres  in  all  did 
he  sell?  Ans.  1218.288  acres. 

11.  A.  bought  from  one  dealer  36.75  tons  of  coal ;  from  a 
second,  25.3165  tons;  from  a  third,  3.75605  tons;  and  from 
a  fourth,  47.62  tons.    How  much  did  he  purchase  from  all? 

A71S.  113.44255  tons. 

12.  At  Pittsburgh,  the  rain-fall  in  Spring  is  9.38  inches ; 
in  Summer,  9.87  inches;  in  Autumn,  8.23  inches;  and  in 
Winter,  7.48  inches.     What  is  the  raiu-fall  for  the  year? 

Ans.  34.96  inches. 

13.  In  Washington  City  the  rain-fall  in  the  Spring  is 
10.45  inches ;  in  Summer,  10.53  inches ;  in  Autumn,  10.15 
inches;  in  Winter,  10.07  inches.  Kequired  the  rain-fall  for 
the  year.  Ans.  41.20  inches. 


ADDITION. 

(14) 
413.403 

(15) 
432.5931 

(16) 
.48321 

4.12356 

96.845 

.6973 

4193.48 

9.38 

.831 

95.631 

.9 

.952 

4.1284 

1.93 

.7345 

786.47 

75.125 

.12345 

5497.23596 

616.7731 

3.82146 

(17) 
179.0009 

(18) 

.  987.789 

(19) 
5.5 

268.008 

98.89 

44.44 

357.09 

9.9 

333.333 

46,9 

89.98 

66.66 

3. 

789.987 

7.7 

853.9999 

1976.546 

457.633 

89 


20.  What  is  the  sum  of  three  tenths,  twenty-five  hun- 
dredths, nine  thousand,  five  millionths,  forty-five  and 
eighteen  ten-thousandths ?  Ans. 

21.  Each  of  the  two  sides  of  a  rectangular  field  is  215.125 
rods  long ;  and  each  of  the  ends  is  124.75  rods  wide.  How 
many  rods  around  the  field  ?  A7is.  679.75  rods. 

22.  Add  75  ten-millionth s,  42  ten-thousandths,  37  hun- 
dredths, .427,  57.3004,  and  .044.  Ans.  58.1456075. 

23.  What  is  the  sum  of  47  ten-thousandths,  15  hun- 
dredths, 8  tenths,  492  millionths,  130  ten-thousandths,  and 
5678  ten-millionths?  Ans.  .9687598. 

24.  What  is  the  sum  of  41.371  -f-  2.29  +  73.402  +  1.729? 

25.  What  is  the  sum  of  86.005  +  4.00003  +  2.000098  ? 


SECTION    III 


mmmTmMmTn&m  % 


127.   Example.  — From  246.7235  take  27.8726. 


SOLUTION. 

246.7235 

27.8726 

218.8509 
liundredtlis ; 


Explanation. — We  write  tlie  subtrahend  so  that  its 
units  stand  under  the  units  of  the  same  order  in  the 
minuend.  Then  subtract  exactly  as  in  whole  numbers. 
6  ten-thousandths  from  15  ten-thousandths  —  9  ten- 
thousandths  ;  3  thousandths  from  3  thousandths  =  0 
thousandths ;  7  hundredths  from  12  hundredths  =  5 
9  tenths  from  17  tenths  =  8  tenths ;  8  ones  from  16  ones 
=  8  ones ;  3  tens  from  4  tens  =  1  ten ;  0  hundreds  from  2  hundreds 
=  3  hundreds. 

Then  we  place  the  decimal  point  between  the  ones  and  tenths,  or 
directly  under  the  decimal  point  in  the  subtrahend. 

128.  EuLE.  —  Write  ike  suhtraliend  so  that  its  units  shall 
stand  under  the  units  of  the  same  order  in  the  minuend. 

Subtract  as  in  integers ;  and  ivrite  the  decimal  point  in 
the  remainder  directly  under  the  decimal  point  in  the  sub- 
trahend. 

PRO  BTj  EMS. 


(1) 

2.7536 

2.4578 

.2958 

From 


(2) 
.3748 
.2697 
.1051 


(3) 
247.36 
25.9734 


(4) 
34.4897 
21.93 


seventy-five  thousandths, 

90 


221.3866  12.5597 

seventy  and  five  tenths  take  seven   hundred 

Am.  69.725. 


SUBTRACTION.  91 

6.  From  eighteen  and  forty-two  ten-thousandths  take 
eighteen  hundred  forty-two  ten-thousandths.     A71S.  17.8Ji. 

7.  From  $.875  take  $.3125.  A7is.  $.5625. 

8.  From  five  thousand  seventeen  millionths  take  .0000097. 

Ans.  .0044073. 

9.  From  Pittsburgh  to  Piiiladelphia  by  the  P.  R.  R.  the 
distance  is  352.2  miles;  from  Pittsburgh  to  Altoona  it  is 
116.7  miles.  What  is  the  distance  from  Altoona  to  Phila- 
delphia? Ans.  235.5  miles. 

10.  By  the  P.,  W.  &  B.  R.  R.  the  distance  from  Pittsburgh 
to  Washington  is  327  miles;  from  Pittsburgh  to  Cumber- 
land, 149.5  miles.  What  is  the  distance  from  Cumberland 
to  Washington?  Ans.  177.5  miles. 

11.  The  average  length  of  the  school  term  in  Pennsyl- 
vania in  1863  was  5.433  months,  in  1869  it  was  0.04  months. 
What  was  the  increase  ?  Ans. 

12.  In  1863,  the  average  cost  per  month  for  each  pupil 
in  the  common  schools  of  Pennsylvania  was  $.49 ;  in  1869, 
it  was  $.97.    What  was  the  increase  ?  A7is.  $.48. 

13.  At  Fort  Humboldt,  California,  the  rain-fall  in  Winter 
is  15.03  inches;  in  Summer,  1.18  inches.  How  much 
greater  is  it  in  Winter  than  in  Summer  ?    Ans.  13.85  inches. 

14.  Chicago  is  .0293317  days  west  of  Washington,  D.  C, 
and  the  Allegheny  Observatory  is  .0082252  days  west  of  the 
same  place.  How  far  is  Chicago  w^est  of  the  Allegheny 
Observatory  ?  A7is.  .0211065  days. 

15.  One  hectare  of  land  equals  2.471  acres.  How  much 
land  have  I  left  after  selling  1  hectare  from  5.5  acres? 

Ans.  3.029  acres. 

16.  Subtract  four  hundred  twenty-five  ten -thousandths 
from  four  hundred  and  twenty-five  ten-thousandths. 

Ans.  399.96. 


SECTION    IV. 


^V=: 


MUIL,TIi^IL,I(€JATIOjM 


^:^ 


139.  Multiplication  of  Decimals  is  tlie  same  as  Multipli- 
cation of  Integers  with  the  exception  of  locating  the  Decimal 
Point. 

Example.  — Multiply  13.25  by  9.27. 

Explanation.— 13.25  multiplied  by 
7  =r  92.75  "  A  "  (89).  But  as  our  multi- 
plier is  .07,  we  have  to  divide  this  product 
by  100,  which  is  done  by  removing  the 
point  two  places  to  the  left — "  B." 

13.25  multiplied  by  2  =  20.50  "A." 
But  as  our  multiplier  is  .2,  we  have  to 
divide  this  product  by  10,  which  is  done 
by  removing  the  point  one  place  to  the 
left-"B." 
13.25  multiplied  by  9  =  119.25  "A,"  "  B." 
Notice  First.— That  with  each  stu;ceeding  figure  of  the  multiplier 
there  is  1  figure  less  to  cut  off.  So  if  each  partial  jwoduct  is  carried  one 
place  further  to  the  left  than  the  preceding  it  will  bring  the  decimal 
point  in  it  exactly  under  the  one  above,  and  consequently  like  orders  will 
stand  in  the  same  columns  in  the  partial  products. 

Notice  Second.— 7^  ''A"  and  "  B"  the  9  ones  of  the  3d' partial 
product  is  obtained  by  multiplying  the  ones  of  the  multiplicand  by  the 
ones  of  the  multiplier ;  and  the  decimal  point  is  placed  at  the  right  of 
ones.    Hence, 


A 

B 

13.25 

13.25 

927 

9.27 

92.75 

.9275 

2  6.50 

2.650 

119.2  5 

119.25 

122.8275 

130.    Tlie  product  of  ones  multiplied  ly  ones  is  ones. 

92 


M  U  L  T  I  P  L  1  C  A  T  I  O  i^ 


93 


131.    The  number  of  decimal  figures  in  a  product  must 
equal  the  number  of  decimal  figures  in  the  factors. 

-Multiply  3.04  by  3.2. 

Explanation. — The  product  of  the  3  ones  of  the  mul- 
tiplicand and  the  3  ones  of  the  multiplier  is  the  9  of  the 
second  partial  product.     Therefore,  that  figure  is  ones, 
and  the  9  of  the  required  product  is  ones  also. 
Notice,  that  there  are  three  decimals  in  the  factors  and 
Q  «  OQ         three  in  the  product  (131). 


133.  Rule. —  Write  the  multiplier  under  the  multipli- 
cand. 

Multiply  as  in  whole  numbers,  and  point  off  from  the 
right  of  the  product  as  many  places  for  decimals  as  there 
are  decimal  places  in  the  factors.  If  there  are  not  enough 
places  in  the  product,  make  a  sufficient  number  by  prefixing 
ciphers. 


FROBT.EM8. 

(1)             (3)             (3) 

(4) 

(5) 

.075          .084          6.375 

63.75 

637.5 

.07              .3                .8 

.08 

5.1000 

.008 

.00525        .0252        5.1000 

5.1000 

Multiply               Ans. 

Multiply 

6. 

.05  by  500.            25. 

11.  7.94  by  4.5. 

7. 

.35  by  225.       78.75. 

12.  12.65  by 

.24. 

8. 

.144  by  12.       1.728. 

13.  48.125  bj 

^  1.08. 

9. 

.0003  by  500.        .15. 

14.  .625  by  64. 

10. 

.75  by  1000.        750. 

15.  61.76  by 

.0071. 

16.  Multiply  3.7267  by  10;  by  100;  by  1000. 


94  DECIMALS. 

SOLUTION.  Explanation. — Since  moving 

3  7267  X        10  37  2G7  ^  figure   one   place   to  the   left 

3*7267  X     100  -  S72  67  i^^^^^^^s  its  value  ten-fold,  and 

d./.50/   X     lUU_^/x5.0/  gij^p3  tl^g    relative    position    of 

3.7207  X  1000  =  3726.7  the  figure  and  decimal  point  is 

the  same  whether  we  move  the 
figure  one  place  to  the  left  or  the  decimal  point  one  place  to  the  right, 
we  multiply  3.7267  by  10  by  moving  the  decimal  point  one  place  to 
the  right. 

By  the  same  process  of  reasoning  we  multiply  by  100  by  removing 
the  decimal  point  two  places  to  the  right,  etc.,  etc. 

17.  What  cost  17.35  tons  of  hay  at  $17,375  per  ton  ? 

18.  What  cost  13.75  dozen  eggs  at  1.1875  per  dozen  ? 

19.  What  cost  18.75  pounds  of  butter  at  $.1875  per 
pound  ? 

20.  Multiply  three  hundred  and  three  thousand  one  hun- 
dred forty-one  ten-thousandths  by  twenty-four  and  three 
thousand  five  hundred  sixty-one  hundred-thousandths. 

Ans.  7218.232585101. 

21.  There  are  16.5  feet  in  1  rod.  How  many  feet  in 
9.8  rods?  Ans.  161.7  ft. 

22.  There  are  5.5  yards  in  1  rod.  How  many  yards  in 
80  rods?  Ans.  440  yards. 

23.  Bought  143.5  acres  of  land  at  $75.25  per  acre.  What 
did  the  whole  cost?  Ans.  $10798.375. 

24.  AVhat  is  the  product  of  1.25  multiplied  by  800  ? 

25.  What  cost  4000.7  bushels  com  at  $.75  a  bushel? 


(26.) 

(27.) 

(28.) 

(29.) 

91.24 

40.083 

.125 

1.25 

1.15 

1.3 

.125 

52.1 

30.  (4.5  X  1.8)  +  (12.5  X  .8)  =  ?  Ans.  18.1. 

31.  (9.2  -  .52)  X  (123  —  114.32)  =  ?        Ans.  75.3424. 

32.  (683.21-1-316.79  -  999.83)  x  .17  =  ?      Ans.  .0289. 


SECTION    V. 


■^ 


©l^ISI^M  I 


h^ ^ 


^r 


3  cents  are  contained  in  6  cents  2  times. 

3  ones  are  contained  in  6  ones  2  times. 

3  tenths  are  contained  in  (5  tenths  2  times. 

3  hundredths  are  contained  in  6  hundredths  2  times. 
3  thousandths  are  contained  in  6  thousandths  2  times. 


Hence, 


133.   When  the  divisor  and  dividend  are  like  nnmhers, 
or  are  of  the  same  order  of  units,  the  quotient  is  ones. 

Example.— Divide  37.18  by  1.43. 


SOLUTION. 

1.43  )  37.18  (  26 
28  6 

8  58 
8  58 


Explanation. — The  right-hand  figures 
of  the  divisor  and  dividend  are  of  the 
same  order  of  units  (hundredths) ;  hence 
the  quotient,  26.  is  an  integer. 


P  BOB  LEMS. 


1.  Divide  5.85  by  .65. 

2.  Divide  292.5  by  22.5. 

3.  Divide  59.163  by  .481. 

4.  Divide  4.732  by  .004. 

5.  Divide  59.994  by  3.333. 


?.  9. 

Ans.  13. 

Ans,  123. 

Ans.  1183. 

Ans.  18. 


134.  The  dividend  must  contain  at  least  as  many  deci' 
mal  figures  as  the  divisor. 


DECIMALS, 


Example.— Divide  30  by  1.25. 


SOLUTION. 

1.25  )  30.00  (  24 
25  0 
5  00 
5  00 


Explanation. — The  right-hand  figure 
of  the  divisor  is  hundredths. 

We  annex  two  decimal  ciphers  to  the 
dividend,  thus  making  its  right-hand  fig- 
ure hundredths,  and  obtain  ones  for  the 
quotient  (134). 


PROBLEMS. 


1.  279  -^  4.5  =  ? 

2.  444  -^  9.25  =  ? 


Ans.  62. 

Ans.  48. 


3.  21  -j-  .375  =  ?      Ans.  56. 

4.  91.2  -^  .95  =  ?     Ans.  96. 


5.  5.1  -T-  .075  =  ?    Ans,  68. 

One-half  of  6  cents  is  3  cents:         6  cents -^2  =  ^  cents. 

One-half  of  6  ones  is  3  ones :  6  ones  -j-  3  =  2  ones. 

One-half  of  6  tenths  is  3  tenths :     6  tenths  -^3  =  2  tenths. 

One-half  of  6  hundredtlis  is  3  hundredths  :    .06  -4-  3  =  .02. 
One-half  of  6  thousandths  is  3  thousandths.        3  ).006 

.002 

135.   When  the  divisor  is  an  integer,  the  quotient  will  be 
of  the  same  orders  of  units  as  the  dividend. 

Example.— Divide  .4375  by  35. 

solution. 
35  )  .4375  ( .0125 
35 
87 
70_ 
175 
175 


Explanation.— The  divisor  35  is  an 
integer.  And  since  the  dividend  is 
tenths,  hundredths,  thousandths,  and 
ten-thousandths,  the  quotient  consists 
of  the  same  orders  of  units  (135). 


PBOBLEMS. 

1.  Divide  .096  by  16.  Ans.  .006. 

2.  Divide  .0162  by  18.  Ans.  .0009. 


Division.  97 

3.  Divide  .0425  by  25.  Ans.  .0017. 

4.  Divide  .00144  by  12.  Ans.  .00012. 

5.  Divide  1728.12  by  12.  Ans,  144.01. 

136.  The  quotient  contains  as  many  decimal  figures  as 
the  tiumber  of  those  in  the  dividend  exceeds  those  in  the 
divisor. 

Example.— Divide  4.6875  by  3.125. 

soLtrriON.  Explanation. — ^The  right-hand  fig- 

3.125  )  4.68715  (  1.5  ^^e  of  the  divisor  is  thousandths,  and 

0  -1 25  til©  corresponding  order  of  the  divi- 
dend  is  7  (thousandths).    Hence,  the 

1  5625  quotient  of  4.687  divided  by  3.125  is 
15625  an  Luteger  ( 1 33). 

Placing  the  decimal  point  after  the 
quotient  figure,  1,  and  completing  the  division,  the  quotient  obtained 
is  1.5,  which  contains  one  decimal  place  (136). 

The  dotted  vertical  line  in  the  solution  shows  what  figures  of  the 
dividend  are  used  to  obtain  the  integral  part  of  the  quotient. 

When  decimal  ciphers  are  annexed  to  form  partial  dividends,  they 
are  counted  as  decimal  figures  of  the  dividend. 

PBOBLEHrS. 

1.  Divide  30.688  by  22.4.  Ans.  1.37. 

2.  Divide  5.7531  by  45.3.  Ans.  .127. 

3.  Divide  92.92261  by  9.119.  Ans.  10.19. 

4.  Divide  .0060858  by  .483.  Ans.  .0126. 

5.  Divide  7.6875  by  .041.  Ans.  187.5. 

6.  Divide  64.86316  by  561.1.  Ans.  .1156.     * 

7.  Divide  .0000798  by  .0042.  Ans.  .019. 

8.  Divide  .6336  by  .132.  Ans.  4.8. 

9.  Divide  20.04  by  .04.  Ans.  501. 
10.  Divide  522,1054  by  42.31.  Ans.  12.34 

7 


98  DECIMALS. 

137.  The  right-hand  figure  of  any  remainder  after  divi- 
sion is  always  of  the  same  order  of  units  as  the  last  figure  of 
the  dividend  used  to  oMain  it. 

Example. — Divide  15.5  by  1.3. 

SOLUTION.  Explanation.— In  the  solution  of  the 

1.3  )  15.5  (  11  following  problems  it  is  necessary  that 

23  the  quotient  shall  consisi  of  an  integer 

only. 

^  ^  Since  the  25  in  the  accompanying  solu- 

1  3  tion  19  tenths,  the  12  is  also  tenths,  and 

-1  o  T>      '  :,  the  true  remainder  is  1.2. 

!./«  Remainder. 

1.  Into  how  many  building  lots,  each  containing  .48  of 
an  acre,  can  8  acres  of  land  be  divided  ? 

Ans.  16  lots,  and  .32  of  an  acre  left. 

2.  How  many  ice-pitchers,  each  weighing  2.13  pounds, 
can  a  manufacturer  make  from  15  pounds  of  silver  ? 

Ans,  7  pitchers,  with  .09  of  a  pound  left. 

138.  EuLE. — Divide  as  in  whole  numbers,  and  point  off 
from  the  right  of  the  quotient  as  matiy  places  for  decimals 
as  the  numher  of  decimal  places  in  the  dividend  exceed  the 
number  in  the  divisor. 

If  there  are  not  so  many  places  in  the  quotient,  supply  the 
deficiency  hy  prefixing  ciphers. 

If  in  the  course  of  division  ciphers  have  been  annexed  to 
the  dividend,  regard  each  cipher  annexed  as  a  decimal  place. 

PMOBZJBMS. 

1.  Divide  .7  by  10.  Ans.  .07. 

2.  Divide  .875'  by  100.  Ans.  .00875. 


DIVISION. 

3. 

Divide  .56 'by  .007. 

Ans.  80. 

4. 

Divide  172.8  by  .012. 

Ans.  14400. 

5. 

Divide  11.7  by  .0009. 

Ans.  13000. 

6. 

Divide  14.4  by  .00006. 

Ans.  240000. 

7. 

Divide  5.85  by  .65. 

Ans,  9 

8. 

Divide  45.3  by  .15. 

Ans.  302, 

9. 

Divide  100  by  .01. 

Ans.  10000. 

10. 

Divide  72  by  .18. 

Ans.  400. 

11. 

Divide  .72  by  18. 

A71S.  .04. 

12. 

Divide  96  by  .016. 

Ans.  6000. 

13. 

Divide  .096  by  16. 

Ans.  .006. 

14. 

Divide  .2  by  1.4. 

Ans.  .142857  +  . 

99 


Note. — The  sign  +  after  the  answer  to  Problem  14  mdicates  that 
the  division  is  not  exact. 


15.   Divide  63.75  by  100  ;  by  1000  ;  by  10000. 


SOLUTION. 

100  =.6375 
63.75-^    1000  =  .06375 


63.75 


63.75 


10000  =  .006375 


Explanation. — Moving  a  fig- 
ure one  place  to  the  right  or 
moving  the  decimal   point  one 
place  to  the  left,  divides  by  10  ; 
two  places  divides  by  100 ;  &c., 
&c. 
Moving  the  decimal  point  of  63.75  two  places  to  the  left  is  the  same 
as  moving  each   figure  two  places  to  the  right,  and  consequently 
divides  each  by  100,  giving  .6375. 

Moving  the  point  three  places  to  the  left,  divides  by  1000,  &c.,  &c. 


16.  Divide  5386.5    by  1000 

17.  Divide  8382.55  by  10000 


18.  Divide    8.75    by  100000 

19.  Divide  10.075  by  1000000 


20.  A  merchant  buys  10000  yards  of  calico  for  $645. 
What  does  it  cost  him  per  yard  ? 

21.  Divide  5  tenths  by  1  trilhon.     Ans.  .0000000000005. 


100  DECIMALS. 

22.  Divide  56.285  by  6.3.  Ans.  8.9341 +  . 

23.  Divide  108.029  by  7.2.  Ans.  15.004  +  . 

24.  Divide  7500  by  one  thousand  eight  hundred  seventy- 
five  millionths.  Ans.  4000000. 

25.  Divide  .75  by  1875.  Ans.  .0004. 

26.  If  2.75  pounds  of  butter  cost  $1.03125,  what  is  the 
price  per  pound  ?  Ans.  1.375. 

27.  If  $8264.50  pay  137.19025  tax,  what  is  the  tax  on  one 
dollar?  Ans.  $.0045. 

28.  Gave  $23.4375  for  75  pounds  of  coffee.  What  was 
the  price  per  pound  ?  Ans.  $.3126. 

29.  The  mail  train  runs  from  Philadelphia  to  Pittsburgh, 
a  distance  of  355.2  miles,  in  17.3  hours.  What  is  the  rate 
per  hour?  Ans.  20.53+  miles. 

30.  The  "  limited  mail "  runs  the  same  distance  in  10.3 
hours;  what  is  the  rate  per  hour  ?       Ans.  34.48+  miles. 

31.  The  fast  mail  runs  from  Pittsburgh  to  Chicago,  a 
distance  of  469  miles,  in  13.084  hours.  What  is  the  rate 
per  hour?  Ans.  35.845+  miles. 

32.  If  2,3  yards  of  cloth  make  1  coat,  how  many  coats 
will  150  yards  make  ?  Ans.  65  coats,  .5  yards  left. 

33.  How  much  muslin  at  $.165  a  yard  can  be  bought  for 
$732.60?  Ans.  4440  yards. 

34.  At  $18.75  each,  how  many  washstands  can  be  bought 
for  $506.25  ?  Ans.  27  washstands. 

35.  A  man  walks  65.625  miles  in  17.5  hours.  How  many 
miles,  on  an  average,  does  he  walk  each  hour  ? 

Ans.  3.75  miles. 


'Y^ 


SECTION    VI 


UiNlliTE^Dj  SJrAfTiES;  MiOjNlEYr 


*-t5X5^(>'eM  •"■ 


139.  United    States   Money,    sometimes    called 
Federal  Money,  consists  of  dollars,  dimes,  cents,  and  mills. 

140.  TABLE. 


10  mills  (m)  =  1  cent . . .  c. 

1000  mills. 

10  cents         =  1  dime. .  .  d. 

%1  = 

100  cents. 

10  dimes        =  1  dollar.  .  % 

10  dimes. 

141.  The  Unit  of  U.  S.  Money  is  the  Dollar, 

Since  the  dollar  is  the  unit  of  U.  S.  money,  dimes,  cents,  and  mills 
are  respectively  tenths,  hundredths,  and  thmtsandth^  of  the  unit. 

142.  Dollars  should  be  written  as  integers,  with  the  sign 
(I)  prefixed ;  and  dimes,  cents,  and  mills  as  decimals.    Thus, 

9  dollars  2  dimes  5  cents  6  mills,  are  written  $9,256. 

The  denomination  dime  is  not  regarded  in  business  operations, 
dimes  being  tens  of  cents.  Thus,  $9,256  is  read  9  dollars  S5  cents 
6  mills,  instead  of  9  dollars  ^  dimes  5  cents  and  6  mills. 

143.  Since  the  two  places,  tenths  and  hundredths,  are 
appropriated  to  cents,  when  the  number  of  cents  is  less  than 

10  write  a  cipher  in  the  place  of  tenths.     Thus,  5  cents  is 

written  $.05. 

In   business  transactions,  if  the  mills  in  the  final  result  are  5  or 

more  than  5,  they  are  considered  a  cent ;  if  less  than  5,  they  are  not 

regarded.     Thus,  $3,186  would  be  called  13.19  ;  $9,424,  $9  42  ;  $.375, 

$.38 ;  &c. 

101 


102  DECIMALS. 


EXERCISES. 


1.  Read  14.12,  $.18,  $3.05,  $10.10,  $5,153. 

2.  Read  $.04,  $24,125,  $125,003,  $99,995. 

3.  Read  $500.50,  $13,033,  $.007,  $199,053. 

4.  Read  $1,005,  $100,009,  $4,875,  $164,664. 

5.  Read  $1000.40,  $89,091,  $75.75,  $77,777. 

6.  Write  15  cents  ;  5  cents ;  9  mills. 

7.  Write  13  dollars  9  cents  ;  5  dollars  80  cents. 

8.  Write  100  dollars  40  cents  3  mills. 

9.  Write  30  cents  6  mills  ;  8  cents  7  mills. 

"    10.  Write  5  dollars  50  cents  ;  1000  dollars  7  cents. 

144.    Decimal  parts  of  a  dollar  less  than   mills    are 
expressed  as  decimals  of  a  mill. 


$  .0003 

is 

2  tenths  of  a  mill. 

$  .00025 

is 

25  hundredths  of  a  mill. 

$  .000125 

is 

125  thousandths  of  a  mill. 

$  .0052 

is 

5  and  2  tenth  mills. 

$  .3552 

is 

35  cents  5  and  2  tenth  mills. 

$4.1007 

is 

4  dollars  10  cents  and  7  tenth  mills. 

EXEItCI  S  E  S  , 

11.  Read  $.054,  $111.0006,  $.49075,  $55.5555. 

12.  Read  $99.43753,  $100.00001,  $.7283,  $99.0888. 

13.  Read  $.40404,  $.25025,  $9.0999,  $10.7558. 

14.  Write  45  cents  7  and  13  hundredth  mills. 

15.  Write  3  dollars  and  3  tenth  mills. 

16.  Write  95  dollars  37  cents  and  5  tenth  mills. 

17.  Write  100  dollars  10  cents  4  and  9  tenth  mills. 

18.  Write  11  cents  3  and  125  thousandth  mills. 

19.  Write  5  cents  and  3125  ten-thousandth  miUs. 

20.  Write  44  dollars  44  cents  4  and  4  tenth  mills. 


UKITED     STATES     MONEY 


103 


[^!i 


^I>jI>)IT'IOjM 


'<^J^f^ 

145.  Since  the  dollar  is  the  unit  of  TJ.  S.  Money  (141), 
and  cents,  mills,  and  parts  of  a  mill  are  decimals  of  a  dollar, 
United  States  Money  is  added  in  the  same  manner  as  other 
decimals. 

Examples.— What  is  the  sum  of  $100.50,  $4,075,  $.43125, 
and  $900.90  ? 

SOLUTION. 

$100.50 
4.075 
.43125 

900.90 


$1005.90625 

(1) 

$481.26 

32.75 

1.25 

421.16 


Explanation. — We  write  the  numbers  so  that 
units  of  the  same  name  stand  in  the  same  column ; 
as,  cents  in  the  column  of  cents,  mills  in  the  column 
of  mills,  etc. 

We  then  add  the  parts  and  place  the  decimal 
point  in  the  sum,  as  in  Addition  of  Decimals,  not 
forgetting  to  prefix  the  sign  ($  )  to  the  result. 


PR  OBLEM, 


$936.42 


(3) 
$593,715 
26.43 
5.421 
18.62 

$644,186 


(3) 
$4.22 
5.031 
8.2004 
9.136 
$26.5874 


(4) 

$.41 
.092 
.03 
.468 

$1. 


5.  Mr.  B.  pays  $91.75  State  taxes;  $321.16  County  taxes; 
$421.13  School  taxes;  $375.08  Borough  taxes;  and  $39.17 
Poor  tax.    What  is  the  total  amount  of  his  taxes  ? 

Ans.  $1248.29. 


104  DECIMALS. 

6.  A  mercliant  deposits  in  bank  on  Jan.  1,  $300 ;  Jan.  3, 
$1716.50;  Jan.  4,  $5217.45;  Jan.  9,  $16714.25;  Jan.  10, 
$516.08;  and  Jan.  11, 17060.37.  How  much  in  all  has  he 
deposited?  Ans.  $31524.65. 

7.  A  owes  B  the  following  sums:  $3.75,  $8.25,  $9.62, 
$20.31,  $17.42,  $16.73,  $24.40,  $120.75,  $16.35,  $115.41, 
$25.64,  $71.22,  $41.78,  $168.37,  $415.93.  How  much  does 
A  owe  B?  Ans.  $1075.93. 

8.  From  the  Savings'  Bank,  C  drew  out  the  following 
sums:  $210,  $35,  $48,  $16,  $25,  $45,  $24,  $16,  $13,  $75,  $84, 
$7,  and  $16.13.    How  much  did  he  draw  out  altogether? 

Ans.  $614.13. 

9.  A  merchant  "checked  out"  of  his  bank  $215,  $3125.75, 
$21620,  $17540,  $2063.74,  $216.14,  $25750.75,  $301.06, 
$51.22,  and  $1617.01.     How  much  in  all  did  he  check  out  ? 

Ans.  $72500.67. 

10.  A  man  after  "checking  out'  $947,  found  that  he  had 
remaining  in  bank  $369.89.    How  much  had  he  at  first? 

Ans.  $1316.89. 

11.  After  "checking  out"  $30.43,  $49.34,  $9,  $25,  a  grocer 
found  that  he  had  $711.84  remaining  in  bank.  How  much 
had  he  at  first  ?  Ans.  $825.61. 

12.  A  merchant  bought  of  a  jobber,  prints  amounting  to 
$3520.50;  mush ns,  $3762.75;  linens,  $1733;  silks,  $2375.16; 
woolen  goods,  $7675.87.     How  much  is  the  bill  ? 

Ans.  $19067.28. 

13.  A  man  bought  a  cow  for  30  dollars,  a  horse  for  104 
dollars  60  cents,  and  a  wagon  for  85  dollars  40  cents.  How 
much  did  all  cost  him  ?  Ans.  $220. 

14.  Each  pupil  will  prepare  an  original  problem  for  class 
exercise. 


UNITED     STATES     MONEY. 


105 


mmmTRMmwiBm  ii 


146,    United  States  Money  is  subtracted  in  the  same 
manner  as  other  decimals  (145). 

Example.— From  $98.45  take  $49,375. 


SOLUTION. 

$98.45 
49.375 
$49,075 


(1) 

$.36 

.19 


Explanation.  —  We  write  dollars  under  dollars, 
cents  under  cents,  and  mills  under  mills. 

We  then  subtract,  and  place  the  decimal  point  in 
tLie  remainder  as  in  Subtraction  of  Decimals,  not  for- 
getting to  prefix  the  sign  ( $ )  to  the  result. 


PROBLEMS. 


(2) 
$47.21 

8.564 


(3) 
$275,386 

85.722 


(4) 
$40,009 
21.4304 


$.17  $38,646  $189,664  $18.5786 

5.  A  merchant  sold  on  Saturday,  goods  amounting  to 
$2565.75  ;  on  Monday,  goods  amounting  to  $1075.23.  What 
was  the  dijfference  in  the  two  days'  sale  ?      Ans.  $1490.52. 

6.  If  a  man's  property  is  valued  at  $15725.50,  and  his 
debts  at  $6837.37,  what  is  he  worth  ?  Ans.  $8888.13. 

7.  If  a  man's  property  is  valued  at  $20569,  and  his  debts 
at  $30880,  what  excess  of  debt  has  he  ?  Ans.  $10311. 

8.  I  deposited  in  bank  $1840,  and  drew  out  $475.50.  How 
much  had  I  left  ?  Ans.  $1364.50. 

9.  A  man  has  property  worth  $10104,  and  owes  debts  to 
the  amount  of  $7426.25.  When  he  pays  his  debts,  how 
much  will  be  left  ?  Ans.  $2677.75. 


106  DECIMALS. 

10.  A  man  having  $100000,  gave  away  $1473.72.  How 
much  had  he  left  ?  Ans.  $98526.28. 

11.  A  merchant  owing  $35542.75,  paid  $22560.25.  How 
much  does  he  still  owe?  Ans.  $12982.50. 

12.  A  man  having  $8795.15,  gave  $2309.95  of  it  for  a 
store.     How  much  money  had  he  remaining  ? 

Ans.  $6485.20. 

13.  A  man  bought  a  span  of  horses  for  $537,  and  a  yoke 
of  oxen  for  $297.50.  How  much  more  did  he  pay  for  the 
horses  than  for  the  oxen  ?  Ans.  $239.50. 

14.  A  man  gave  me  his  note  for  $240,  and  he  has  since 
paid  all  but  $40.50  of  it.  How  much  has  he  paid  on  the 
note?  Ans.  $199.50. 

15.  A  bank-clerk  earned  $150  a  month,  and  his  expenses 
were  $67.43.     How  much  did  he  save?         Ans.  $82.57. 

16.  I  paid  $25000  for  a  piece  of  property  and  afterward 
sold  it  for  $19400.75.   What  was  my  loss  ?  Ans.  $5699,25. 

17.  From  $11.05  take  $9,258.  Ans.  $1,792. 

18.  From  $.001  take  $.00095.  Ans.  $.00005. 

19.  From  $10  take  $5.75.  Ans.  $4.25. 

20.  From  $5  take  $1.85.  Ans.  $3.15. 

21.  I  pay  $5268  for  property  which  I  sell  for  $4319.67. 
What  is  my  loss  ?  Ans.  $948.33. 

22.  From  9  dollars  subtract  8  dollars  99  cents  9  mills. 

Ans.  $.001. 

23.  Subtract  99  dollars  99  cents  9  mills  from  $100,875. 

Ans.  $.876. 

24.  From  1  dollar  subtract  $.0555.  A^is.  $.9445. 

25.  Each  pupil  will  prepare  an  original  problem  for  class 
exercise. 


UNITED     STATES     MON^EY 


107 


^-5)«(5- 


m  mmi/F'i^iji(^;MT'iom 


^ 


147.   United  States  Money  is  multiplied  in  the  same 
manner  as  other  decimals  (145). 

Example.— Multiply  $13.75  by  4.5. 

SOLUTION. 


$13.75 
4.5 

6875 
5500 
$61,875 


Explanation. — We  write  the  multiplier  under 
the  multiplicand. 

We  then  multiply,  and  place  the  decimal  point  in 
the  product,  as  in  Multiplication  of  Decimals,  not 
forgetting  to  prefix  the  sign  ($)  to  the  result. 


1.  Multiply  $137.57  by  7. 

2.  Multiply  $37,354  by  6. 

3.  Multiply  $12,225  by  8. 

4.  Multiply  $.347  by  9. 

5.  Multiply  $2575  by  .04. 

6.  Multiply  $33.75  by  .008. 

7.  Multiply  $.40  by  500. 

8.  Multiply  $57.27  by  504. 

9.  Multiply  $.875  by  56. 

10.  Multiply  $100  by  .00001. 

11.  If  a  boy  goes  to   the  opera  at 
evening,  how  much  will  it  cost  him  for 


Ans.  $962.99. 

Ans.  $224,124. 

Ans,  $97.80. 

A71S.  $3,123. 

Ans.  $103. 

Ans.  $.27. 

Ans.  $200. 

Ans,  $28864.08. 

Ans.  $49. 

Ans.  $.001. 

a  cost  of  $1.25  an 

20  evenings  ? 

Ans.  $25. 


108  DECIMALS. 

12.  A  gentleman's  expenses  for  one  year  were  $3450.75. 
At  the  same  rate  what  would  they  be  for  25  years  ? 

Ans.  $86268.75. 

13.  What  cost  450  tons  of  iron  at  $45.75  per  ton? 

Ans.  $20587.50. 

14.  What  cost  37  tons  of  steel  at  $63.50  per  ton  ? 

Ans.  $2349.50. 

15.  What  cost  372  yards  of  muslin  at  11  cents  per  yard  ? 

Ans.  $40.92. 

16.  What  cost  568  yards  of  calico  at  6  cents  per  yard  ? 

Ans.  $34.08. 

17.  A  street-car  conductor  receives  $1.75  per  day.     How 
much  does  he  make  in  one  year  of  365  days  ? 

Ans.  $638.75. 

18.  An    institution    spends  in   advertising  $565.34  per 
year.     At  this  rate  how  much  would  it  spend  in  15  years  ? 

Ans.  $8480.10. 

19.  If  a  boy  places  in  a  Savings'  Bank  $1.15  per  week, 
how  much  will  he  deposit  in  1  year  or  52  weeks  ? 

Ans.  $59.80. 

20.  A  farmer  sold   493   pounds  of  wool  at  $.875   per 
pound.     How  much  did  he  receive  for  it  ?    Ans.  $431.38. 

21.  A  tinner  worked  5.5  days  at  $1,375.  per  day.     How 
much  did  he  earn  ?  Ans.  $7.56. 

22.  What  cost  47  bushels  of  potatoes  at  $.655  per  bushel  ? 

Ans.  $30.79. 

23.  What  cost  35  yards  of  carpeting  at  $2,375  per  yard  ? 

Ans.  $83.13. 

24.  What  cost  77  bushels  apples  at  $.875   per  bushel  ? 
At  $.625  per  bushel?  Ans.  1st.,  $67.38;  2d.,  $48.13. 

25.  Each  pupil  will  prepare  an  original  problem  for  class 
exercise. 


UNITED     STATES     MONEY. 


109 


Jj^ 


MFWIMJi&jm 


148.   United  States  Money  is  divided  in  the  same  manner 
as  other  decimals  (145). 

Example.— Divide  $61,875  by  4.5. 


SOLUTION. 

4.5  )  $61,875  (  $13.75 
45^_ 
16  8 
13  5 

3  37 

315 


225 
225 


Explanation.— We  write  the  di- 
visor at  the  left  of  the  dividend. 

We  then  divide,  and  place  the 
decimal  point  in  the  quotient,  as  in 
Division  of  Decimals,  not  forgetting 
to  prefix  the  sign  ($)  to  the  result, 
when  the  divisor  is  an  abstract  num- 
ber (102). 


PBOBIjEMS, 

1.  Divide  $962.99  by  7.  Ans,  1137.57. 

2.  Divide  $224,124  by  6.  Ans.  $37,354. 

3.  Divide  $97.80  by  8.  Ans.  $12,225. 

4.  Divide  $3,123  by  9.  Ans.  $.347. 

5.  Divide  $103  by  .04.  Ans.  $2575. 

6.  Divide  $.27  by  .008.  A^is.  $33.75. 

7.  Divide  $200  by  500.  Ans.  $.40. 

8.  Divide  $28864.08  by  504.  Ans.  $57.27. 

9.  Divide  $49  by  56.  Ans.  $.875. 
10.  Divide  $.001  by  .00001.  Ans.  $100. 


110  DECIMALS. 

11.  Divide  13.123  by  $.347.  Ans.  9. 

12.  Divide  $103  by  $2575.  Ans.  .04. 

13.  Divide  $.27  by  $33.75.  Ans,  .008. 

14.  Divide  $200  by  $.40.  A71S.  500. 

15.  Divide  $28864.08  by  $57.27.  Ans.  504. 

16.  Divide  $49  by  $.875.  Ans.  56. 

17.  Divide  $.001  by  $100.  Ans.  .00001. 

18.  If  shovels  are  worth  $1.15  apiece,  how  many  can  be 
had  for  $41.40?  Atis.  36  shovels. 

19.  At  $.875  per  bushel,  how  many  bushels  of  corn  can 
be  bought  for  $83,125  ?  Ans.  95  bushels. 

20.  A  merchant  sold  2000  barrels  of  flour  for  $14750. 
How  much  did  he  get  per  barrel?  Ans.  $7,375. 

21.  A  puddler's  income  one  year  of  313  working  days  was 
$2582.25.    What  was  the  average  per  day?      Ans.  $8.25. 

22.  A  family  of  seven  persons  spent  in  one  year  of  365 
days  $1642.50.    What  was  the  average  per  day  ? 

Ans.  $4.50. 

23.  Bought  456  bushels  of  bituminous  coal  for  $50.16. 
How  much  was  that  per  bushel  ?  Ans.  $.11. 

24.  At  6  cents  a  yard  how  many  yards  of  calico  can  be 
purchased  for  $34.08  ?  Ans.  568  yards. 

25.  If  a  street-car  conductor  earns  $1.75  per  day,  how 
long  will  he  be  in  earning  $638.75  ?  Ans,  365  days. 

26.  If  a  boy  deposits  $1.15  per  week  in  a  Savings'  Bank,  in 
how  many  weeks  will  he  have  deposited  $59.80  ? 

Ans.  52  weeks. 

27.  A  hvery  man  received  $14208  for  boarding  37  horses, 
12  months.  How  much  did  he  charge  a  month  for  each 
horse?  Ans.  $32. 

28.  Each  pupil  will  prepare  an  original  problem  for  class 
exercise. 


OUTLINE    OF    FRACTIONS. 


150.  FBACTIONAJL  UNIT. 


p2 


181. 
182. 
183. 
184. 
185. 
186. 
I  187. 


151.  TERMS. 


CLASSIFICA- 
TION. 


PRINCIPLES. 


PREPARATORY.  ^ 


j    152. 
]   153. 


Denominator. 
Numerator. 


As  to  Value. 


As  to  Form. 


li 


As  to  Fractional 
Unit. 


154.  Proper. 

155.  Improper, 

156.  Simple. 

157.  Complex. 
5  8.  Compound. 

159.  Similar. 

160.  Dissimilar. 


161.  First.— Multiplied. 

162.  Second.— Divided. 

163.  Ttiird.—Not  changed. 

«  f  166.  Z)mswsj  167.  Prime. 
k^  b  J         or i^actor*.  (  168.  Composite. 
2t  1   169.  Common  Divisor  or  Factor. 

fe  i.  170.  Greatest  Common  Divisor. 


1'71.  Multi 


lulti-^  1 
pies.  \  1 


72.  Commwi  Mvltiples. 

73.  Least  Com.  Mult. 


CASES. 


V  174.  Cancellation. 

f  175.  jr.  Integers  or  Mixed  Num- 
bers to  Improper  Fractions. 

176.  JTJ.  Improper  to  Integers  or 

Mixed  Numbers. 

177.  III.    Integers    to  Higher  or 

Lotver  Terms. 

178.  IV.  To  Lowest  Terms. 

179.  V.  Dissimilar  to  Similar. 

180.  VI.  Dissimilar  to  Least  Sim- 

ilar. 


ADDITION. 

SUBTRACTION. 

MULTIPLICATION. 

DIVISION. 

CONVERSE  REDUCTION, 

ALIQUOT  PARTS. 

BILLS. 


Ill 


6    SIXTHS^  5,   seveMTHS^  8    EIGHTHS^ 

iiiiiiilgj^^  iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiniiiiit!! ' n!!!!," ""'"if 


CHAPTER    III. 

SECTION    I. 


(2i 


(5? 


-^®^ 


FB'ACT'10)MS 


149.  A  Fraction  is  a  number  expressing  one  or  more 
of  the  equal  parts  of  a  unit. 

150.  A  Fractional  Unit  is  one  of  the  equal  parts 
into  which  a  unit  is  divided. 


112 


lifUMERATIOK     AlfD     NOTATIOlf.  113 


The  number  of  these  parts  indicates  their  name.     Thus, 
One  divided  into  two  equal  parts  =  2  halves.     One  half  is  written  |, 
One  divided  into  three  equal  parts  =  3  thirds.      Two  thirds   are 
One  divided  mio  four  equal  parts  =  A  fourths.  Three  fourths" 
One  divided  into  fioe  equal  parts  =  5  fifths.      Four  fifths     " 
One  divided  into  six  equal  parts  =  6  sixths.      Five  sixths     " 
One  divided  into  seven  equal  parts  =  7  sevenths.  Six  sevenths  " 
One  divided  into  eight  equal  parts  =  8  eighths.    Seven  eighths 
One  divided  into  nin£,  equal  parts  =  9  ninths.     Eight  ninths  " 
One  divided  into  ten  equal  parts  =  10  tenths.    Nine  tenths    "  ^^^or  .9. 

g 

When  we  express  nine  tenths  thus,  Jq,  it  is  called  a  Common  Frac- 
tion ;  but  when  thus,  .9,  it  is  called  a  Decimal  Fraction.     Hence, 

With  reference  to  the  mode  of  expressing  them.  Fractions  are  of 
two  kinds,  namely :  Common  and  Decimal.    (See  page  81.) 

A  fraction  is  expressed  by  two  numbers,  one  written  under  the 
other,  with  a  horizontal  line  between  them. 

151.  The  Terms  of  a  fraction  are  the  two  numbers 
used  to  express  it,  and  are  called  the  Denominator  and  the 
Numerator. 

152.  The  Denominator  is  that  term  of  the  fraction 
which  indicates  the  number  of  equal  parts  into  which  the 
unit  is  divided,  and  is  written  below  the  horizontal  hne. 

153.  The  Numerator  is  that  term  which  indicates 
the  number  of  equal  parts  taken,  and  is  written  above  the 
horizontal  line.  Thus,  in  the  fraction  f ,  the  3  is  the  denomi- 
nator, and  signifies  that  a  unit  is  divided  into  three  equal 
parts ;  the  2  is  the  numerator,  and  signifies  that  two  of  these 
parts  (thirds)  are  taken,  or  expressed. 

Fractions  are  read  by  first  pronouncing  the  number  in  the  numera- 
tor, and  then  the  parts  of  the  unit  expressed  by  the  denominator. 
Thus  ^  is  one  half ;  |,  two  thirds;  ^,  four  twenty-firsts;  ^,  seven 
thirty-seconds. 
8 


114 


FKACTIOKS. 


Read 

2*         2f         sy         a>        sf 

T?         T>         "!>         i>       Tf 
ity     TT?     A^     yf  >    4o> 

Write 

1.  Four  ninths. 

2.  Six  twenty-firsts. 

3.  Five  eighteenths. 

4.  Seven  thirty-seconds. 

5.  Eight  twelfths. 

6.  Nine  twenty-fifths. 

7.  Two  nineteenths. 

8.  Eight  elevenths. 

9.  Eleven  fifteenths. 


JEx^ncisjss. 


h     h     h     h     h     h     -V, 
m   A,    A,   A    H,   «,    «, 

I  10.  Nine  one-hundred-thirds. 

11.  Two  three-hundred-fortieths. 

12.  Five  two-hundredths. 

13.  One  four-hundred-thirteenth. 

14.  Three  three-hundred-firsts. 

15.  Seventy  eightieths. 

16.  Nineteen  ninetieths. 

17.  Four  ten-thousandths. 

18.  Two  hundred-thousandths. 


In  reference  to  their  value,  fractions  are  of  two  kinds,  namely; 
Proper  and  Improper. 

154.  A  Proper  Fraction  is  one  whose  numerator  is 

less  than  its  denominator ;  as,  ^,  f . 

Such  a  fraction  is  called  proper  because  its  value  is  less  than  a  unit, 
and  is,  therefore,  properly  expressed  in  &  fractional  form. 

155.  An  Improper  Fraction  is  one  whose  numer- 
ator equals  or  exceeds  its  denominator;  as,  f,  ^. 

Such  a  fraction  is  called  improper  because  its  value  equals  or  exceeds 
a  unit,  and,  therefore,  the  proper  expression  for  its  value  would  be  an 
integer  or  a  mixed  number. 

EXEJtClSES, 

Of  the  following  fractions,  name  the  class  to  which  each 
belongs. 

T>        f  ?        f >         T>       "^?        T»      TO?      "TT"?      ^^9     TU9 
A»      lT»     "V*      TtS*      Tt>      4T>      T*>        M>      TSTir* 


KUMERATIOK     AND     l^OTATION.  115 

A  Mixed  Number  is  one  expressed  by  an  integer  and  a  fraction ;  as, 

In  reading  a  mixed  number,  and  belongs  between  the  integer  and 
fraction.     Thus,  4|  is  read  4  and  ^. 

In  reference  to  their  form,  fractions  are  of  three  kinds,  namely. 
Simple,  Complex,  and  Compound. 

156.  A  Simple  Fraction  is  one  whose  terms  are 

whole  numbers ;  as,  |,  f. 

157.  A  Conipleoc  Fraction  is  one  which  has  a  frac- 
tion  in  one  or  both  of  its  terms ;  as,  |,  -^. 

158.  A  Compound  Fraction  is  a  fraction  of  a 
fraction ;  as,  \  of  ^,  which  may  also  be  written  thus,  |  x  J. 

In  reference  to  their  fractional  unit,  fractions  are  of  two  kinds, 
namely :  Similar  and  Dissimilar. 

159.  Similar  Fractions  are  those  that  have  like 
fractional  units;  as,  |,  |^;  -^^j,  /j,  v'V- 

160.  Dissimilaf*  Fractions  are  those  that  have 
unlike  fractional  units ;  as,  f ,  J ;  J,  |,  J. 

EXERCISES. 

Write  in  separate  groups  the  similar  fractions  found  in 
the  following: 

T>        6'        ^^      Tfy      h      T?       iSf      TT>      T7'        "8?        T?      IT'      T> 

*,   A,   h   A.   I>   f.     I.    «.    I.    «.   i    ?.    4. 
a,   I,    f.     4.    A.  A.  ¥,    V.    tt>   ¥.  «.  ^,  ¥• 

If  we  multiply  the  numerator  of  f  by  2,  we  obtain  |.  The  frac- 
tional unit  in  |  and  |  is  the  same,  |.  |  has  3  ^imes  as  many  fractional 
units  as  f .     |  ^  2  _  4      Again, 

If  we  divide  the  denominator  of  f  by  2,  we  obtain  f .  The  number 
of  fractional  units  in  |  and  f  is  the  same,  2.     But  the  value  of  the 


116  FKACTIOi^S. 

fractional  unit  :^  is  3  times  the  value  of  the  fractional  unit  \.    f  ^.  2  =  f . 
Hence, 

161.  A  fraction  is  multiplied  hy  multiplying  its  nu- 
merator or  dividing  its  denominator,  iy  a  number  greater 
than  unity. 

If  we  divide  the  numerator  of  f  by  2,  we  obtain  \.  The  fractional 
unit  in  f  and  \  is  the  same,  \.  \  has  only  one  half  as  many  fractional 
units  as  |.    |  -*-  »  =  ^.     Again, 

If  we  multiply  the  denominator  of  |  by  2,  we  obtain  |.  The  number 
of  fractional  units  in  f  and  |  is  the  same.  But  the  valae  of  the  frac- 
tional unit  \  is  only  one  half  the  value  of  the  fractional  unit  \. 
f  X  2  =  |.     Hence, 

162.  A  fraction  is  divided  hy  dividing  its  numerator 
or  multiplying  its  denominator,  ly  a  number  greater  than 
unity. 

If  we  multiply  both  terms  of  \  by  2,  we  obtain  |.  The  number  of 
fractional  units  in  |  is  twice  as  many  as  in  J,  but  the  xoHue  of  each 
fractional  unit  is  only  one  half  as  much.  That  is,  the  increase  in  the 
number  of  parts  taken  equals  the  decrease  in  the  size  of  the  parts. 
ixi=|.     Again, 

If  we  divide  both  terms  of  |  by  2  we  obtain  \.  The  number  of  frac- 
tional units  in  \  is  only  one  half  as  many  as  in  f ,  but  the  value  of  the 
fractit'-nal  units  is  twice  as  much.  That  is,  the  decrease  in  the  number 
of  parts  taken  equals  the  increase  in  the  size  of  the  parts.  1 1:  |  =  ^. 
Hence, 

163.  The  value  of  a  fraction  is  not  changed  hy 

either  multiplying  or  dividing  hoth  terms  hy  the  same  number. 

Fractions  primarily  arise  from  performing  the  operation  of  division, 
when  the  division  is  not  exact  (99,  1). 

That  of  which  the  fraction  expresses  a  part  is  called  the  Unit  of 
the  Fraction  ;  as  in  the  expression  $|,  one  dollar  is  the  Unit  of  the 
Fraction. 

The  Unit  of  the  Fraction  is  not  always  a  single  thing ;  it  may  be  a 
collection  of  things  taken  as  a  whole  ;  as  in  the  expression  |  of  fifty 
men,  fifty  men  is  the  unit  of  the  fraction. 


SECTION    II. 


I  mmmmm'Enmm  I 


-^--'**^®(^^'-->- 


164.  deduction  of  Fi'actions  is  changing  them 
into  other  equivalent  expressions.  It  also  includes  the 
changing  of  whole  and  mixed  numbers  to  the  form  of  a 
fraction. 

Before  proceeding  with  Reduction  of  Fractions  it  is  necessary  to 
understand  something  of  Measures  and  Multiples. 

165.  A  Divisor^  or  3feasurej  of  a  number  is  a 
number  which  will  divide  that  number  without  a  remain- 
der.    Thus,  3  is  a  divisor,  or  measure,  of  6. 

"  Measure"  is  here  used  in  the  limited  sense  of  exact  divisor  (99, 
1).  Any  number  may  be  used  as  the  measure  of  any  other  number  of 
the  same  kind  (191,  259). 

166.  The  Divisors  or  Factors  of  a  number  are  the 
numbers  which  multiplied  together  will  produce  it.  Thus, 
2  and  3  are  factors  of  6 ;  2,  6,  and  5,  of  60. 

An  Even  Number  is  one  that  is  exactly  divisible  by  2.  The  left- 
hand-page  numbers  of  this  book  are  even. 

An  Odd  Number  is  one  that  is  not  exactly  divisible  by  2.  The  right- 
hand-page  numbers  of  this  book  are  odd. 

A  Prime  Number  is  one  that  can  not  be  separated  into  integral  fac- 
tors ;  as  2,  3.  5,  7, 11.  18. 

The  number  1  is  not  regarded  as  a  factor. 

167.  The  I^Htne  Factors  or  Frime  Divisors  of 

a  number  are  the  prime  numbers  which  multiplied  together 

117 


118  FKACTIONS. 

will  produce  it.  Thus,  2,  6,  and  5,  or  2  and  30,  or  5  and  12 
are  factors  of  60 ;  but  2,  2,  3,  and  5  are  the  prime  factors 
of  60,  because  they  are  the  prime  numbers  whose  product 
is  60. 

When  the  numbers  are  small, we  can  find  the  prime  factors  by  in- 
spection ;  but  when  they  are  large,  we  have  generally  to  obtain  them 
by  trial. 

In  making  trial  a  knowledge  of  the  following  facta  may  be  of  ser- 
vice : 

First. — Any  number  ending  in  0,  2,  4,  6,  or  8  has  2  for 
an  exact  divisor. 

Second. — Any  number  the  sum  of  ivhose  digits  is  divisi- 
ble by  3,  has  3  for  a  divisor. 

Third. — Any  number  ending  in  5,  or  0,  has  5  for  an  exact 
divisor. 

Example. — Find  the  prime  factors  of  2310. 

SOLUTION.  Explanation. — From  the  1st  Fact,  we 

n  N  nqi  rv  see  that  3  is  an  exact  divisor  of  2310. 

'- From  the  2d  Fact,  we  see  that  3  is  an 

3  )  1155  exact  divisor  of  the  quotient  1155. 

5  \  335  From  the  3d  Fact,  we  see  that  5  is  an 

.     -,  exact  divisor  of  the  quotient  385. 

f  None  of  these  Facts  aid  us  in  obtaining 

11  the  next  factor ;  but  since  7  is  a  prime 

number  and  a  divisor  of  77,  we  divide  by 
Ans,  2x3x5x7x  11.    it  and  obtain  11,  a  prime  number,  for  the 
quotient. 
The  product  of  the  divisors  2,  3,  5,  7,  and  the  last  quotient,  11,  all 
of  which  are  prime  numbers,  is  2310.    Hence  tli?y  are  the  prime  fac- 
tors of  that  number  (167). 

KuLE. — Divide  by  any  prime  number  except  1  that  is  an 
exact  divisor  ;  divide  the  quotient  in  the  same  manner  ;  and 
so  proceed  until  the  quotient  is  a  prime  number. 

The  several  divisors  and  the  last  quotient  are  the  prime 
factors. 


EEDUCTIOlf, 


119 


EXERCISES. 

Find  the  prime  factors  of 


1. 

6. 

Ans.  2,  3. 

11. 

32. 

Ans.  2,  2,  2,  2,  2. 

2. 

8. 

Ans.  2,  2,  2. 

12. 

34. 

Ans.  2,  17. 

3. 

12. 

Ans.  2,  2,  3. 

13. 

96. 

Ans.  2,  2,  2,  2,  2,  3. 

4. 

16. 

^7^5.  2,  2,  2,  2. 

14. 

132. 

Ans.  2,  2,  3, 11, 

5. 

18. 

^W5.  2,  3,  3. 

15. 

144. 

Ans.  2,  2,  2,  2,  3,  3. 

6. 

20. 

^/i5.  2,  2,  5. 

16. 

385. 

Ans.  5,  7, 11. 

Z 

24. 

^/i5.  2,  2,  2,  3. 

17. 

1001. 

Ans.  7,  11,  13. 

8. 

27. 

^ws.  3,  3,  3. 

18. 

4862. 

Ans.  2,  11,  13, 17. 

9. 

28. 

Ans.  2,  2,  7. 

19. 

8008. 

Ans.  2,  2,  2,  7,  11,  13, 

10. 

30. 

Ans,  2,  3,  5. 

20. 

2310. 

Ans.  2,  3,  5,  7, 11 

A  composite  number  is  one  tliat  can  be  separated  into  inte- 
gral factors ;  as  4,  6,  8,  9,  10, 13. 

168.  The  Composite  Factors  or  Divisors  of  a 

number  are  the  composite  numbers  which  multiphed  together 
will  produce  it.  Thus,  4  and  15  are  the  composite  factors  of 
60,  because  they  are  the  composite  numbers  whose  product 
is  60.  . 

If  a  number  is  an  exact  divisor  of  two  or  more  numbers,  it  is  said 
to  be  a  divisor,  or  measure,  common  to  them. 

169.  A  Common  Divisor  or  Measure  of  two  or 

more  numbers  is  an  exact  divisor  of  each  of  them.  Thus,  2 
is  a  divisor  common  to  2,  4,  6,  8,  10,  &c.,  and  is  called  a 
common  divisor. 

Numbers  are  prime  to  each  other  when  1  is  their  only  exact  divisor. 

170.  The  Greatest  Common  Divisor  or  Meas- 
ure of  two  or  more  numbers  is  the  greatest  number  that  is 
an  exact  divisor  of  each  of  them.    Thus,  6  is  the  greatest 


120  REACTIONS. 

number  that  will  exactly  divide  12  and  18,  hence  it  is  their 
greatest  common  divisor. 

Example.  —  Find  the  greatest  common  divisor  of  105 
and  175. 

FIRST  METHOD. 

SOLUTION.  Explanation.  —  Resolving 

105  =  3  X  5  X  7  l^*h  numbers  into  their  prime 

^^^5  =  5  X  5  X  "7  factors,  we  find  5  and  7  to  be 

TT.i  c        w        n^     A  tlie    only   factors    common    to 

In  both  are  5  X  7  =  35,  Ans.  .  *i         x.        mr  •        -,    . 

>         "  both  numbers.    Their  product, 

5x7,  therefore,  is  the  greatest  common  divisor. 

SECOJ^D   METHOD. 

BOLunoN.  Explanation.  — 105  is  the  greatest 

105)  175  (1  measure  of  itself.    If  it  also  measured 

-J  ^-  175  it  would  be  the  greatest  divisor  of 

both  numbers. 

70)  105  (1  But  it  does  not  measure  175,  because 

YO  70  remains.     That  is,  175  =  105  +  70. 

Now,  as  70  is  the  greatest  measure  of 

Ans.    35 )  70  ( 2       itself,  if  it  also  measures    105    it  will 
70  measure  175. 

But  it  does  not  measure  105,  because 
35  remains.  TMfet  is,  if  175  =  105  +  70,  then  175  =  35  +  70  +  70. 
Now,  as  35  is  the  greatest  measure  of  itself,  if  it  measures  70  it  will 
measure  105  (which  =  35  +  70),  and  175  (which  =  35  +  70  +  70). 

Since  35  does  measure  70,  it  will  also  measure  105  and  175.  It 
must  be  the  greatest  common  divisor,  because  it  is  tbe  greatest  num- 
ber contained  in  105  and  in  the  difference  between  105  and  175. 

KuLE  I. — Find  the  product  of  all  the  prime  factors  com- 
mon to  the  given  numlers. 

Or,  II. — Divide  the  greater  number  hy  the  less,  and  if  there 
is  a  remainder,  divide  the  preceding  divisor  ly  it.  So  con- 
tinue dividing  the  last  divisor  hy  the  last  remainder,  till 
nothing  remains.  The  last  divisor  will  le  the  greatest  com- 
mon divisor  of  the  two  numbers. 


EEDUCTIOK.  121 

WJien  iliere  are  more  than  two  numbers,  first  find  the 
greatest  common  divisor  of  tivo  of  them,  then  of  that  com- 
mon divisor  and  one  of  the  other  numbers,  and  so  on  till 
every  number  has  been  used.  The  last  common  divisor  is  the 
greatest  common  divisor  of  all  the  numbers. 


EX  ERC  I  S  JES, 

Find  the  greatest  common  divisor 

1.   Of  48  and  72. 

Ans. 

24. 

2.   Of  36  and  54. 

Ans. 

18. 

3.   Of  60  and  270. 

Ans. 

30. 

4.   Of  180  and  300. 

Ans. 

60. 

5.   Of  32  and  56. 

Ans. 

8. 

6.   Of  91  and  143. 

Ans. 

13. 

7.   Of  20,  50,  and  70. 

Ans. 

10. 

8.    Of  30,  42,  and  63. 

Ans. 

3. 

9.   Of  120,  210,  and  345. 

Ans. 

15. 

10.    Of  154,  210,  and  287. 

Ans. 

7. 

11.   Of  330,495,  and  660. 

Ans. 

165. 

171.  A  Multiple  of  a  number  is  any  number  which 
is  divisible  by  it.     Thus,  15  is  a  multiple  of  5. 

172.  A  Common  Multiple  of  two  or  more  numbers 
is  a  number  which  is  divisible  by  each  of  them.  Thus,  12 
is  a  multiple  common  to  2,  3,  4,  and  6. 

173.  The  Least  Common  Multiple  of  two  or  more 
numbers  is  the  least  number  that  is  divisible  by  each  of 
them.  Thus,  12  is  the  least  common  multiple  of  2,  3,  4, 
and  6,  because  it  is  the  least  number  that  each  will  exactly 
divide. 

Example. —  Find  the  least  common  multiple  of  6,  9, 
and  12. 


I22  fEACTIOJS^S. 


FIRST  METHOD. 
SOLUTION.  Explanation.  —  Resolv- 

n  2    V   3  ^^^  ^®  numbers  into  their 

prime  factors,  we  see  that 

^  ^=^  ^    y^  *^  the  answer    must    contain 

12  =  3x3x3  2x2x3,  because  12  = 

2x2x3x3=  36,  Ans.     3x2x3. 

-  The  answer    must   also 

contain  3x3,  because  9  =  3x3.    But,  as  there  is  already  one  3  in 
2  X  2  X  3,  we  need  only  use  another  3  making  2x2x3x3. 

The  answer  must  also  contain  2x3,  because  6  =  2x3.  But  in 
2x2x3x3  we  already  have  2x3. 

Therefore  2x2x3x3  contains  all  the  prime  factors  of  all  the 
numbers,  and  none  of  them  more  times  than  is  necessary  to  contair/ 
every  number.  Hence  2x2x3x3  =  36  is  the  least  commor 
multiple  of  6,  9,  and  12. 

Since  in  the  above  Example  6  is  a  factor  of  12,  it  might  have  beei> 
omitted  in  the  solution,  and  the  same  result  obtained.    Thus, 

0 


9 

= 

3 

x 

3 

12 

= 

2 

X 

2 

X 

3 

2 

X 

2 

X 

3 

X  3  =  36,  Ans. 


SECOJ^D    METHOD. 

SOLUTION.  Explanation.  —  We   take   the 

2  )   6,  9,  12  factor  2  out  of  6  and  12,  and  then 

the  numbers  become  3,  9,  and  6. 

^  )_A_?i ?  Again  we  take  the  factor  3  out, 

13      2  ^^^  *^^y  become  1,  3,  and  2. 

Ov,  Qv,  Qv,  o        oa     A^c.  Now,  we  cannot  take  out   any 

2x0x3x2  =  36,  Ans,      .,  c   ^     r       ^  o      ^ 

other  common  factor  from  1,  3,  and 

2.  Hence  we  have  rejected  all  factors  which  are  not  necessary  to  con- 
tain 6,  9,  and  12. 

Those  which  are  retained,  namely,  2,  3,  3,  and  2,  must  be  necessary 
to  contain  those  numbers. 

Hence,  their  product,  namely,  2x3x3x2  =  36,  must  be  the 
least  number  that  will  contain  them. 


REDUCTION.  •      123 

Omitting  the  '  6"  as  suggested  in  the  First  Method,  and  we  have  : 
3)  0,  9,  12 

3,     4,  or  3  X  3  X  4  =  36,  Ans. 

Rule  I. — Resolve  the  numbers  into  their  prime  factors, 
and  multiply  together  all  those  of  the  largest  number,  and  such 
of  the  others  as  are  not  found  in  the  largest  number  ;  the 
product  is  the  least  common  multiple. 

Rule  II. —  Write  the  numbers  in  a  line,  and  divide  by 
any  prime  number  that  is  contained  exactly  in  two  or  more 
of  them.  Write  the  quotients  and  undivided  tiumbers  in  a  line 
below.  Divide  these  in  the  same  manner.  Continue  thus  till 
no  number  greater  than  1  is  exactly  contained  in  any  two  of 
the  numbers.  TJien  multi2oly  all  the  divisors  and  remaining 
numbers  together  ;  the  product  is  the  least  common  multiple. 

Wlien  numbers  are  prime  to  each  other,  they  have  no  com- 
mon factor  to  be  rejected,  and  their  least  common  multiple  is 
their  product. 

EXEItCISES. 

Find  the  least  common  multiple 

1.  Of  10,  15,  and  20.  Ans.  60. 

2.  Of  12,  18,  and  24.  Ans.  72. 

3.  Of  14,  35,  and  56.  Ans.  280. 

4.  Of  16,  24,  36,  and  60.  Ans.  720. 

5.  Of  20,  30,  50,  and  75.  Ans.  ZOO. 

6.  Of  44,  66,  88,  and  110.  Ans.  1320. 

7.  Of  65,  78,  104,  and  130.  Ans.  1560. 

8.  Of  48,  80,  120,  and  144.  Ans.  720. 

9.  Of  8,  11,  and  15.  Ans.  1320. 

10.  Of  120,  180,  200,  and  240.  Ans.  3600. 

11.  Of  1,  2,  3,  4,  5,  6,  7,  8,  and  9.  Ans.  2520. 

12.  Of  7,  13,  23,  and  31.  Ans.  64883. 


124       •  FKACTIONS. 

174.  Cancellation  is  the  process  of  ahlreviating  oper- 
ations in  division  by  rejecting  equal  factors  from  the  divisor 
and  dividend. 

The  Sign  of  Cancellation  is  an  oblique  line  drawn  across  a  figure  ;  as 
0,  '^$,  U$,  &c. 

A  factor  is  cancelled  hy  dividing  both  dividend  and  divisor  hy  that 
factor. 

Example. — Divide  4x5x6  by  2x4x5. 

SOLUTION.  Explanation.  —  We    write    the 

3  numbers    that   constitute  tlie  divi- 

^  X  ^  X  P    Q      Ayig  dend  above  a  line,  and  those  that 

^  X  4  X  ^  constitute  the  divisor  below  it. 

Observing  4  and  5  to  be  factors  in 
both  dividend  and  divisor,  we  cancel  them  in  botli. 

Then  observing  that  2  is  a  factor  in  the  2  of  the  divisor  and  in  the 
6  of  the  dividend,  we  cancel  it  out  of  these,  leaving  in  the  dividend 
the  factor  3,  which  is  the  required  quotient. 

Since  dividing  a  number  by  itself  gives  1  for  a  quotient,  if  the  factor 
or  factors  cancelled  equal  the  number  itself,  the  result  will  be  One, 
which  need  not  be  written,  except  in  the  quotient  where  there  are  no  other 
factors. 

EuLE. — Cancel  all  factors  common  to  both  dividend  and 
divisor,  and  divide  the  product  of  the  factors  remaining  in 
the  dividend  by  the  product  of  those  remaining  in  the  divisor. 

EXERCISES. 

1.  Divide  8  x  12  x  15  by  4  x  3  x  12.  Ans.  1 0. 

2.  Divide  6  x  7  x  8  by  3  x  7  x  4.  Ans.  4. 

3.  Divide  9x5x11  by  11x9x5.  Ans.  1. 

4.  Divide  42  x  12  x  18  by  18  x  36  x  7.  Ans.  2. 

5.  Divide  144x360  by  6x8x9x20.  Ayis.  6. 

6.  Divide  8  xl2  x27  by  4  x3  x  12.  Ans.  18. 

7.  Divide  20  x  32  x  35  by  4  x  5  x  16.  Ans.  70. 


EEDUCTION.  125 


MISC  ELIjAN  EOU  S     P  It  O  B  Tj  E  M  S  , 

1.  Find  the  prime  factors  of  325.     Of  1872. 

Ansivers.  5,  5, 13 ;  2,  2,  2,  2,  3,  3, 13. 

2.  Find  all  the  exact  divisors  of  24.     Of  42. 

Answers.  2,  3,  4,  6,  8,  12,  24;  2,  3,  7,  6, 14,  21,  42. 

3.  Find  the  largest  number  that  will  divide  without  a 
remainder  720  and  960.  Ans.  240. 

4.  Find  the  greatest  common  measure  of  2145  and  3471. 

Ans.  39. 

5.  Find  the  least  common  multiple  of  54,  378,  486,  and 
540.  Ans.  34020. 

6.  Find  the  greatest  common  measure  and  the  least  com- 
mon multiple  of  12,  36,  72,  and  144.     Ansivers,  12  ;   144. 

7.  What  must  be  the  width  of  carpeting  to  fit  three  rooms 
18,  21,  and  24  feet  wide  respectively  ?  Ans. 

8.  What  is  the  least  sum  of  money  for  which  I  can  pur- 
chase sheep  at  2,  3,  4,  or  5  dollars  each,  and  just  expend  the 
whole  ?  Ans. 

9.  How  long  is  the  longest  measure  that  will  accurately 
measure  three  pieces  of  cloth,  respectively  27,  63,  and  108 
yards  long  ?  Ans. 

10.  What  is  the  l^st  sum  of  money  for  which  a  man  can 
buy  cattle  at  18,  30,  or  36  dollars  each  ?  How  many  can 
he  buy  at  each  price  ? 

Ans.  $180.    10  at  $18;  6  at  $30,  or  5  at  $36. 

11.  Find  the  value  of  the  following  expression  by  cancel- 
lation :  40  X  54  X  99  X  250  -j-  20  X  27  X  198  X  50.    Ans. 

]  2.  A  man  bought  20  cases  of  muslin,  each  case  contain- 
ing 50  pieces  of  40  yards  each,  at  10  cents  per  yard,  and 
paid  in  coffee,  giving  80  bags  of  200  pounds  each ;  what 
was  the  coffee  per  pound  ?  Ans.  25  cents. 


126  FRACTIONS. 


CASE    I. 
175.   Integers   or  Mixed   Numbers^  to   Im 


proper  Fractions, 


-•>•- 


(i)ral'^:vfGrciXGX  -^ 


-•-♦•• 


^ 


Example. — How  many  thirds  inl?    2?    4?    6? 

Solution. — In  1  there  are  3  thirds  ;  therefore  in  2  there  are  two 
times  3  thirds,  which  are  6  thirds. 

In  4  there  are  4  times  3  thirds,  which  are  13  thirds,  &c. 


PROBT.EMS, 

1.  How  many  fourths  in  3?    8?    6?    5?    9? 

2.  How  many  fifths  in  4  ?    7  ?     11  ?    12  ?     10  ? 

3.  How  many  sixths  in  5?    9?     7?    8?     6? 

4.  How  many  sevenths  in  8?    7?    6?    5?    4? 

5.  How  many  eighths  in  7  ?    5  ?    4  ?    3  ?    1  ? 

In  performing  these  examples  it  will  be  noticed, 
FiKST.  That  we  change  the  form  of  the  numbers,  but  not  their 
rnlue. 

Second.  That  each  whole  number  has  now  the  form  of  a  fraction. 

Example. — How  many  thirds  in  3J^ 

Solution.— Since  in  1  there  are  3  thirds,  in  3|  there  are  3  times 
3  thirds  +  1  third  =  ^/. 

6.  How  many  fifths  in  If  ?    3|  ?     6^  ?  7^  ? 

7.  How  many  eighths  in  3  ?    2|  ?    3^  ?  5|  ? 

8.  How  many  sixths  in  2  ?    4|  ?    3^-  ?  7|  ? 

9.  How  many  fourths  in  4  ?    4^  ?     8f  ?  lOJ  ? 

10.  How  many  sevenths  in  3  ?     5f  ?     6^  ?     8-f-  ? 

11.  How  many  ninths  in  2  ?    2^  ?    5f  ?    6|  ? 


KEDUCTIOK. 


127 


Reduction  of  Fractions  is  changing  their  form  without  alter- 
ing their  value  (104). 


i-lVV^i'ittei^  ^xfercise^-f 


Example  1. — Eeduce  41  to  ninths. 


BOLUTION. 

9 
41 

369 
9  ' 


Ans. 


Explanation. — Since  in  1  there  are  9  ninths,  in 
41  there  are  41  times  9  ninths,  or  369  ninths. 

The  same  answer  may  be  obtained  by  multiplying 
the  whole  number,  41,  by  9,  and  writing  the  result 
over  the  required  denominator. 


Example  2. — Reduce  25^  to  a  fractional  form. 


SOLUTION. 

25H  =  25  +  H- 
12  X  25       11       311 


12 


12  ~  12 ' 


Ans. 


Explanation.— In  1  there  are 
12  twelfths;  therefore  in  25i^ 
there  are  25  times  12  twelfths  + 
11  twelfths  =  ^. 


Hence  the 

Rule. — Multiply  the  given  denominator  by  the  whole 
number.  Add  the  given  numerator ,  if  any,  to  the  product, 
and  write  the  sum  over  the  given  denominator. 


PJROBLEMS 


1.  4  to  twelfths. 

2.  16  to  eighths. 

3.  25  to  thirtieths. 

4.  121  to  140ths. 

5.  132  to  75ths. 


Reduce 
6. 
7. 
8. 
9. 
10. 


167  to  lOOths. 
97  to  30-thirds. 
125  to  25ths. 
16  to  12tbs. 
Ill  to  40-firsts. 


128 


PR  ACTIONS. 


Reduce  to  Improper  Fractions, 


11.  37|. 

12.  341yV 

13.  542|i. 

14.  742|if. 

15.  2147,^. 


16.  17571||. 

17.  13562ff}-. 

18.  26753ff. 

19.  3672^^07. 

20.  7im|. 


CASE  II. 

176.   Improper  Fractions    to    Integers    or 
Mixed   Numbers, 


4-  ©ral^:«fGiici8%X  -^ 


Example  1. — Reduce  ^  to  an  integer. 

Solution. — There  are  3  thirds  in  1.    Since  3  thirds  are  contained  in 
12  thirds  4  times,  12  thirds  =  4. 

Example  2. — Reduce  ^  to  a  mixed  number. 

Solution. — There  s^re  4:  fourths  in  1.    Since  4:  fourths  &re  contained 
in  17  fourths  4  and  1  fourth  times,  17  fourths  =  4^. 

PROBLEMS. 

Reduce 

1.  ¥.  ¥.  ¥>  ¥.  ¥.  and  ¥  to  integers. 

2.  -i^,  ¥  -¥-'  ¥?  and  ^  to  mixed  numbers. 

3.  ¥.  ¥  ¥.  ¥.  ¥.  and  V  to  integers. 

4"  ¥?  ¥>  ¥>  ¥j  and  ^  to  mixed  numbers. 

5.  In  14  o^  a  foot,  how  many  feet  ?    In  ¥  ^^  an  acre, 
how  many  acres  ?    In  $^,  how  many  dollars  ? 

6.  In  -2^  of  a  quart,  how  many  quarts  ?    In  4^  of  a  peck, 
how  many  pecks  ?    In  f  of  a  bushel,  how  many  bushels  ? 


REDUCTION. 

7.  How  many  pints  in  ^  of  a  pint  ?  ^^  of  a  pint  ? 

8.  How  many  days  in  -^  of  a  day  ?  ^  of  a  day 

9.  How  many  miles  in  -^^  of  a  mile  ?  ^  of 

10.  In  ^  of  a  melon,  how  many  melons  ? 

11.  In  ^  of  a  gallon,  how  many  gallons  ? 

12.  ^=?^  =  ?^  =  ?^=?^^=? 


129 


a  mile? 


— l^. 


tlWiitten  ^^erci^G^+ 


-♦•- 


Example  1. — Reduce  -i^  to  an  integer. 

SOLUTION.  Explanation.— Since  12  twelfths  make  1, 

12  )  144  there  are  as  many  I's  in  144  twelfths,  as  12  is 

contained   times  in  144,  which  are  12  times. 

12,  Ans.  Therefore,  ^<  =  12. 

Example  2. — Reduce  f4  to  a  mixed  number. 

ftOT  TnPTO"V 

.  '  Explanation.— Since  there  are  \}  in  1, 

17  )  9b  (  5ff,  Ans.       ^^^^^  ^re  as  many  I's  in  f|,  as  17  is  con- 
85  tained  times  in  96,  which  are  5||  times. 

Therefore,  f«  =  5H. 


11 


Rule. — Divide  the  numerator  ly  the  denominator.  Write 
the  remainder,  if  any,  over  this  denominator  for  the  frac" 
tionalpart  of  the  number. 


PROBLEMS. 

Reduce  to  integers 

1. 

n- 

Ans,  8. 

5.     ^^2^. 

Ans.  91. 

2. 

^m- 

Ans.  12. 

6.    W- 

Ans.  24. 

3. 

-WA 

Ans.  11. 

7.    W. 

Ans.  63 

4. 

¥f- 

Ans.  31. 

8.  m- 

Ans.  1. 

>0 

FEACTIOKS. 

Keduce  to  mixed  numbers 

9. 

w. 

Ans.  7^. 

14.  1^/1. 

^ws.  312|f. 

10. 

^i*^. 

A71S.  52^. 

15.  1^2,41.. 

Ans.  1974JI 

11. 

H-}^' 

Ans.  205f?-. 

16.  i-y^K 

Jws.  754/y. 

12. 

Wf- 

^W5.  8^. 

17.  W. 

Ans.  13|4 

13. 

WA^. 

-4^5.  27,%V 

18.  W. 

^ws.  13H- 

CASE    III. 
mi.  Fractions  to  Higher  or  Lower  Terms. 

A  fraction  is  reduced  to  Higher  Terms  when  the  numerator 
and  Denominator  are  expressed  in  larger  numbers.  Thus  i  =  |,  op 
ijs>  or  iV  &c. 

A  fraction  is  reduced  to  Lower  Terms  when  the  Numerator  and 
Denominator  are  expressed  in  smaller  numbers.    Thus,  ^  =  f ,  or  ^. 

•-♦-• — — 


0i.ar3E:^(3 


Example  1. — In  ^  how  many  ninths  ?    In  |  ?    In  |  ? 

Solution. — Since  in  1  there  are  9  ninths,  in  i  of  1  there  must  be  \ 
of  9  ninths,  or  3  ninths;  in  f  there  must  be  f  of  9  ninths^  or 
6  ninths,  &c. 

PJtOBLEMS. 

1.  In  i  how  many  12ths  ?    Inf?    f?    f?    |?    V?    J? 

2.  In|howmany24tbs?    In|?    |?    |?    |?    V?   ^? 

3.  In^J^^owmany  60ths?    In^f?   H?   if?  A?   A? 

4.  In  I  how  many  25ths  ?    In  ^?    |?    |?    -J? 

5.  In|howmany28ths?    In^?    f?    4?    ^?    J^? 

How  may  the  answers  to  these  questions  be  obtained  ?  Ans.  By 
multiplying  both  terms  of  the  fraction  by  an  integer.     Thus,  \,  |,  &c.. 


REDUCTION.  131 

are  reduced  to  12ths  by  multiplying  both  terms  by  2  ;  -^y*  it»&c.,  are 
reduced  to  GOths  by  multiplying  both  terms  by  5. 

Are  the  values  of  the  fractions  altered  by  this  process?  Why 
not?  (163.) 

Example  2. — How  many  halves  inf?  f?  -j^?  ff? 

Solution.— Since  in  1  there  are  |,  in  ^  there  are  ^  of  |,  or  f .  If 
there  is  |  in  |,  in  |  there  are  as  many  halves  as  2  is  contained  times 
in  4,  which  are  3  times.     Therefore,  |  =  |.    In  like  manner  f  =  |, 

6.  How  many  thirds  in  ^  ?    In  ^  ?  if  ?  f f  ?  ff  ? 

7.  How  many  sixths  in  if  ?     In  H  ?  M  ?  «?  if  ? 

8.  How  many  eighths  in  ^  ?     In  \l  ?  f|  ?  U?  H? 

How  may  the  answers  to  questions  6—8  be  obtained? 
What  effect  on  the  value  of  a  fraction  has  dividing  both  terms  bj 
the  same  number  ?  (163.) 


Example  1. — Reduce  ^  to  a  fraction  whose  denominator 

is  72. 

SOLUTION.  Explanation.— Since  the  denominator,  24,  is 

5    X  3 15       to  be  changed  to  72,  we  must  inquire  by  what 

24  X  3        72       number  24  must  be  multiplied  to  produce  72. 

By  dividing  72  by  24,  we  find  the  number  to 
be  3.  We  multiply  34  by  3,  and  also  the  numerator  by  3,  so  as  not  to 
chanj3re  the  value  of  the  fraction  (163),  and  obtain  ||. 

Example  2.— Reduce  ff  to  24ths. 

SOLUTION.  Explanation.— Since  96  is  to  be  changed  to 

56  ->  4  _  14       24,  we  divide  96  by  34  to  find  the  divisor  that  will 

96  -T-  4        24       effect  the  change.     We  find  it  to  be  4.     Dividing 

96  by  4,  we  must  also  divide  the  numerator  56 

by  4,  in  order  that  the  value  of  the  fraction  may  remain  unchanged 

(163),  and  find  that  f  f  =  ^f . 


132  FEAGTIONS. 

Example  S.—Reduce  J  to  a  fraction  whose  numerator 
is  15. 

SOLUTION.  Explanation.— We  first  divide  15  hj  3  to 

3  X  5  15        find  the  number  by  which  we  must  multiply  the 

4x5        20        given  numerator,  3,  that  the  product  may  be  15. 
We  find  6  to  be  the  number,  by  which  we  mul- 
tiply the  terms,  and  obtain  f  =  ^f . 

Example  4. — Reduce  ff  to  a  fraction  whose  numerator 
is  8. 

Explanation. — We  find  by  dividing  48  by  8 
SOLUTION.  that  6  is  the  number  by  which  the  given  numer- 

48-7-6  S^      ator  must  be  divided  to  produce  the  required 

96-1-6        16      numerator.    Dividing  both  terms  by  6  we  have 
If  =  A- 

Rule. — Multiply  or  divide  both  terms  of  the  fraction  by 
such  a  number  as  will  change  the  giveti  denominator  or  numer- 
ator to  the  required  one. 


PBOBL  EMS, 

Reduce 

1. 

4  to  49ths.         Ans.  ff . 

6. 

H 

to  3ds.          Ans.  f. 

2. 

i  to  72ds.          Ans.  ^. 

7. 

1^ 

to  llths.     Ans.  ^. 

3. 

^  to  576ths.     Ans.  f^. 

8. 

AV 

to  12ths.     Ans.  -^. 

4. 

^  to  1728ths.  Ans.^^. 

9. 

1066 

to  26ths.    Ans.  ^. 

5. 

^^to8784ths.^/i5.-^y. 

10. 

^A 

to  74ths.    Ans.  -f^. 

Reduce 

11.      J    to  numerator  24. 

Ans.  M- 

12.      1    to  numerator  525. 

Ans.  m- 

13.    -^  to  numerator  438. 

Ans.  fjf. 

14.    ^^  to  numerator  273. 

Ans.  m- 

15.    3^5  to  numerator  1243. 

Ans.  ,WiV 

16.     f^  to  numerator  5. 

Ans.  A- 

EEDUCTION.  133 

17.  Reduce  ^i^  to  numerator  12.  Ans,  3^. 

18.  Eediice  y^^  to  numerator  3.  Ans.  ^. 

19.  Reduce  j-||  to  numerator  13.  Ans.  \^. 

20.  Reduce  f|^  to  numerator  14.  Ans,  \4[, 

In  the  preceding  problems,  our  multipliers  and  divisors  are  all  in- 
tegers, but  had  they  been  fractions  the  application  of  the  Itule  would 
have  been  the  same;  for  example,  let  it  be  re(iuired  to  reduce  f  to  a 
fraction  whose  denominator  is  10.  We  divide  10  by  4  to  find  the 
number  by  which  we  must  multiply  both  terms,  and  obtain  ^.  ^^  of 
4  are  10;  ^^  times  3,  or  3  times  1^,  are  -\a  =  72  (176).     Therefore, 

7| 
f  =v^,  a  complex  fraction  (157). 

In  this  Case  we  have  seen  that  a  fraction  is  reduced  to  lower  terms 
by  division  of  both  its  terms  by  a  common  divisor.  It  is  plain,  there- 
fore, that  a  fraction  is  reduced  to  its  lowest  terms  by  the  division  oi 
both  its  terms  by  their  greatest  common  divisor. 

CASE  IV. 
178.  Fractions  to  their  Lowest  Terms. 


©rat^Exfoi^ciXoX  ^^ 


Example. — Reduce  Jf  ho  its  lowest  terms. 

Solution. — Since  6  is  the  greatest  common  divisor  of  12  and  30,  we 
divide  both  terms  by  6  and  have  ^§  =  f .  Therefore,  f  is  ^f  reduced 
to  its  lowest  terms. 

FltO  BZ  E3IS. 

1.  Reduce  to  lowest  terms,  f ,  ^,  ^,  i^,  H,  ^\,  ^. 

2.  Reduce  to  lowest  terms,  ■^,  ff ,  ||,  ff ,  A,  1^. 

3.  Change  H  ^o  its  lowest  terms,     fj.    if.     ||-.    f . 

4.  Change  ^  to  its  lowest  terms,     -ff .    ■^^.    -Vf.    ff- 

5.  Express  with  the  smallest  whole  numbers  the  fractions 

wv,  fi.  m^  m>  and  if. 

A  fraction  is  said  to  be  in  its  Loivest  Terms  when  the  numerator 
and  denominator  are  prime  to  each  other  (169). 


134 


FKACTIONS. 


^-'^Wi'itten  ^^erci^esr-i 


Example. — Reduce  f  j-  to  its  lowest  terms. 


FIRST  SOLUTION. 

45 -^9_5 
54-^9  ~~6 


SECOND  SOLUTION. 


45 

54 


15 

18 


THIRD  SOLUTION. 
5 

5       ^  _5 

6  04    ~  6 


Explanation. — In  the  1st  Solution,  we  first  find  the  greatest  com- 
mon divisor  of  45  and  54  by  (170),  or  by  inspection,  and  divide  by  it 
both  terms,  and  obtain  f  as  the  result. 

In  the  2d  Solution,  we  first  divide  both  terms  by  the  common  factor 
3.  But  as  the  terms  are  not  yet  prime  to  each  other,  we  divide  again 
in  like  manner  by  3.     The  result  is  f ,  in  which  both  terms  are  prime. 

In  the  3d  Solution,  we  simply  canceled  the  factors  common  to  both 
terms.    Hence  the 

Rules. — 1.  Divide  both  terms  ly  their  greatest  common 
divisor;  or, 

2.  Divide  both  terms  by  any  common  factor,  and  the  re- 
sulting terms  in  the  same  manner,  till  they  are  prime ;  or, 

3.  Cancel  all  factors  common  to  both  terms. 


3.  Jf- 

5.  if- 
6.**. 


Ans.  ^. 
Ans.  |. 
Ans.  f. 
Ans.  f. 
Ans.  -J. 
Ans.  -J. 


PJt  O  B  LBMS 

Reduce  to  lowest  terms 


ITHT* 


7. 

9.  in. 

ll'    26  2  5* 


Ans.  -^. 

Ans.  i. 
Ans.  -j^r- 

Ans.  i- 
Ans.  A- 
Ans.  i^. 


13.m- 

15.  m- 

16.  iWr- 

!"•     2  32  3* 


Ans.  -^. 
Ans.  41. 
Ans.  ^\. 
Alls.  -^. 
Ans.  T^. 
Ans.  A-. 


KEDUCTION-  135 

CASE    V. 
179.  Dissimilar  to  Similar  Fractions, 
•4^- 


^Wi^itteH^^rcises^ 


E 

iX  AMPLE   1, 

SOLUTION. 

2 
3 

X 
X 

5 
5 

10 
~15 

4 
5 

X 
X 

3 
3 

12 
~15 

-Reduce  f  and  ^  to  similar  fractions. 

Explanation. — Thirds  cannot  be  reduced  to 
fifths,  nor  fifths  to  thirds,  without  resulting  in 
complex  fractions.     (See  1st  Note,  page  133.) 

But  since  3  times  5  =  15,  we  reduce  f  to 
fifteenths  by  multiplying  its  terms  by  5  ;  and 
since  5  times  3  =  15,  we  reduce  |  to  fifteenths 
by  multiplying  its  terms  by  3.  Hence,  |,  |  = 
\h  if. 

Example  2. — Reduce  \,  {,  and  f  to  similar  fractions. 

solution.  Explanation. — Since  the  product  of  the 

1  V  4  X  6        24  denominators  2,  4,  and  6  =  48,  we  may 

reduce  the  fraction  to  48ths.  This  is  done 
by  multiplying  the  terms  of  each  fraction 

3x2x6        36  by  the  denominators  of  the  other  fractions. 

4x2x6        48  ^^®  terms  of  \  multiplied  by  4  and  6  =  f|; 

those  of  f ,  by  2  and  6  =  ff  ;  and  those  of 

5  X  2  X  4  _  40  f ,  by  2  and  4  =  |§.    Hence,  i,  f ,  |  =  f  |, 

6x2x4      48         11,40. 

Fractions  having  the  same  denominator  are  said  to  have  a  common 
denominator. 

The  common  denominator  of  two  or  more  fractions  is  a  com-tnon 
multiple  of  their  denominators. 

Reducing  "dissimilar  to  similar  fractions"  is  sometimes  called 
reducing  fraMions  to  equivalent  fractions  having  a  common  denominator. 

Rule. — Multiply  loth  terms  of  each  fraction  hy  all  the 
denominators  of  the  other  fractions. 


2x4x6      48 


136 


rUACTIOJ^S. 


1*  hobijEms, 

1.  Reduce  |,  f,  and  |  to  similar  fractions. 

■^ns.  \i,  1^,  if. 

2.  Reduce  j,  |,  and  |  to  similar  fractions. 

^^S'  m,  1%,  *. 

3.  Reduce  |  and  ^  to  similar  fractions.  Ans.  ^,  f|. 

4.  Reduce  f  and  f  to  similar  fractions.  Ans.  ||,  f  J. 

5.  Reduce  |  and  -^  to  similar  fractions.  ^ws.  ||,  |f. 

CASE    VI. 
180.  Dissimilar  to  Least  Similar  JFractions, 


^%ym{iG^;^^e 

(rcisi^ 

1 

£ 
5 

In  examining  the  fractions  f|  and  f f  (answer  to  Prob.  3,  Case  V), 
■we  find  that  we  may  reduce  them  to  lower  terms,  and  still  preserve  a 
common  denominator  ;  thus,  by  dividing  both  terms  of  f f  and  f^,  by 
4,  we  reduce  them  to  ^  and  ^  respectively.  And  since  they  cannot 
be  reduced  to  any  lower  terms  and  at  the  same  time  have  a  common 
denominator,  they  are  said  to  be  reduced  to  least  similar  frac- 
tions. 

The  least  common  denominator  of  two  or  more  fractions  is  the 
least  common  multiple  of  their  denominators. 

Reducing  "  dissimilar  to  least  similar  fractions  "  is  sometimes  called 
redttcing  fractions  to  equivalent  fractions  hairing  a  least  common  denom- 
inator. 

Example. — Reduce  ^^,  -^,  and  -^  to  least  similar  fractions. 

Explanation.  —  We 

BOLTJTION. 

A?  A?  A?  *^®  fractions. 
20, 18,  12,  their  denominators. 
180,  the  L.  C.  M.  of  the  denominators 
9, 10, 15,  L.  0.  M.  -T-  denominators. 


first  find  the  L.  C.  M.  of 
the  denominators  20, 18, 
12,  to  be  180  (173). 

We  now  find  our  mul- 
tipliers to  be  respective- 
ly 9,  10,  15  (page  131, 
Ex.  1^. 


EEDUCTION.  13? 

_3^  X  9    21  Multiplying    both     terms 

20  X  9    "  180  of  ^  by  9,  ^\  by  10,  and  ^ 

^    y^lQ         50  ^7 15,  we  have  ^\\,  ■^%,  and 

18  X  10  ^  180  ^t         .       A       A'   '   ' 

Hence,  to  reduce  dissiuu- 

7    X  15 105  lar  to  least  similar  fractions, 

12  X  15  ~"  l80  we  have  this 

EuLE. — Multiply  loth  terms  of  each  fraction  ly  the  num- 
her  of  times  its  denominator  is  contained  in  the  least  common 
multiple  of  all  the  denominators. 

PROBLEMS. 

Eeduce  to  least  similar  fractions, 

1.  I,  ^\,  and  H-  ■^^«-  tt  41.  H' 

2.  A,  ii,  and  fi-.  Ans.  -^,  ^,  «f 

3.  f,  h  I,  tV  ^^^-  H'  u>  ii  n 

4.    f,  A,  f,  U'  ^^s-  A.  «.  M.  4i- 

5.  i  i  i  i  f  ^'^^^  «. » li^  n^  u 


6.  I,  1,  h  ^.  ^^^-  *V,  tli.  m 

7.  I,  -I,  A,  U'  ^^''  lo-.  H,  f  ^,  U 

8.  V-.  ii  J-  ^^^-  -W.  M,  W 

9.  h  I  f.  ^^5.  ill.  m>  -ih 

10.  \S  ^/,  M.  ^^^-  W'  W.  -If 

11.  4,  i,  f  ^^5.  u,  if,  if, 

13-  A  A.  A-  ^^^-  -^^^  A.  U 

13.  Yjj,  YW'  "^*  Ans.  -ff^,  2^,  -^^ 

14.  A,  I,  A.  ^/2^.    H,  if,  *■ 

15.  i  «,  if.  Ans.  If  If,  if, 

16.  A.  i  A-  -4;?5.  -^,  If,  ^, 

17.  i,  A,  A.  ^^^.  If,  A,  li- 

18-  i,  h  h  h  ^ns.  ff ,  iA,  ^,  If 

19.  i  ih-,  A  ^ns.  A¥f,  i¥A-.  tMt 

20.  i,  i,  i  ^/?5.  A\,  ^^  -tt 


SECTION    III. 


< 


<G)©^ 


/l^niDlTIQlMI 


181. 


i;^. 


0i*aL  3E:^ei^ciXeX 


Example  1. — What  is  the  sum  of  }  and  J  ? 

Solution. — Since  3  and  1  are  4,  dfourtlis  and  1  fourth  must  be  4 
fourtlis  or  1. 

1.  What  is  the  sum  of  land  f?  f  andf  ?  fand}? 

2.  What  is  the  sum  of  f  and  i?  i  and  |  ?  f  and  f  ? 

Example  2. — What  is  the  sum  of  |  and  f  ? 

Solution. — The  least  common  denominator  of  f  and  f  is  85. 
f  =  fi  ;  f  =  it-    Therefore,  the  sum  of  f  and  f  is  the  sum  of  |^  and 

4.  What  is  the  sum  of  f  and  |?  f  and  i?  -^  and  yV? 

Example  3.— Find  the  sum  of  3^  and  4|. 

Solution. — The  sum  of  the  integers  3  and  4  is  7.  The  sum  of  the 
fractions  ^  and  ^  is  f  +  f  =  f ,  or  1^ ;  and  7  + 1^  =  8^.  Therefore,  the 
sum  of  3|  and  4f  is  %\. 

138 


ADDITION.  139 

6.  71+8i  =  ?  9f4-7i  =  ?  m  +  3\  =  ?  5|  +  4|=? 

7. 51+91  =  ?  3i+8i  =  ?  ^-{-^  =  ?  n+H=? 

8.  I  paid  $1^  for  some  steak,  and  8^  for  vegetables.    How 
much  did  I  pay  for  all  ? 

9.  A  man  having  J  of  an  acre  of  ground,  bought  f  of  an 
acre  more.     How  much  had  he  then  ? 

10.  Jennie  gave  l-J-  for  a  slate,  $J  for  a  book,  and  $^  for 
paper.     What  did  she  pay  for  all  ? 

11.  A  boy  gave  $20J  for  a  coat,  $7J  for  a  pair  of  boots,  and 
$3|  for  a  hat.    How  much  money  did  he  expend  ? 

12.  lOOi  +  250|  4-  175|  +  25|  =  ? 


-•-♦►#- 


j'lWi'itte^  ^xevci^e^-i- 


k- 


Example  1.— What  is  the  sum  of  ^,  t^,  and  ^  ? 

Explanation.— The  sum  of  3  units, 

and  5  units,  and  4  -wwi^*  is  13  units. 

_  _i   _    I    _  :^  _.      Therefore,  the  sum  of  3  seventeentJis,  and 

17       17       17        17        5  seventeenths  and  4  seventeenths  is  13 

seventeenths. 

Example  2.— Find  the  sum  of  ^,  |,  and  ^, 

SOLUTION.  Explanation.— The  f rae- 

5         3         7  tions  being   dissimilar,  we 

j^  +  g  +  Jg  ^=  first  reduce    them   to    the 

80        78        91_249_.,      f™"-  '«-=»-»«  ^..  A\ 


208        208        208        208"   ^'''        Since  the  parts  of  these 

similar  fractions  are  all  of 
the  same  kind,  208ths,  and  since  the  numerators  express  the  numbers 
of  these  parts,  we  add  the  similar  fractions  by  adding  their  numer^ 
ators  80,  78,  and  91  ;  and  since  the  fractional  unit  of  these  parts  is  ^, 
we  write  the  denominator  208,  under  the  249,  making  f^f.  Reducing 
1^1  to  a  mixed  number,  we  have  1^^,  Atis. 


140  FRACTIONS. 

Example  3.— Find  the  sum  of  3|,  21f ,  and  14|. 

SOLUTION.  Explanation.— We  first  find 

^i  =    3-4-^=    ^+fi  *li®  sum  of  the  integers  3,  21, 

214  =  21  +  -f  =  21  +  fj  and  14  to  be  38 :  next,  we  find  tlie 

■1  j5  14  4.  4 14  4_  44  s^"^  ^^  *^^®  fractional  parts,  ^,  f > 

**  —        -Mr  —  ^lJt  ^j^^  5  to  be  m     Finally,  we  add 

^"  H"  l^'  together  the  two  sums  38  and 
Or,  38  +  If^  1ft,  and  find  our  entire  sum  to 
Or  39M     ^^  39ft,  the  result  required. 

Rule. — If  the  fractions  have  a  common  denominator, 
write  over  it  the  sum  of  all  the  numerators. 

If  the  fractions  have  not  a  common  denominator,  reduce 
them  to  a  coiimion,  or  to  their  least  common  denominator, 
and  then  write  the  sum  of  the  numerators  over  the  common 
denominator. 

Add  mixed  numbers  hy  finding  the  sum  of  the  integral 
and  fractional  parts  separately,  and  then  adding  their  sums. 


P  B  OBLEM8. 

1.  Find  the  sum  of  ^,  If,  and  Jf  Ans.  2^. 

2.  Find  the  sum  of  ff ,  44?  and  ff-  Ans.  Iff. 

3.  Find  the  sum  of  ^,  -Jf ,  \-^,  and  ^.  Ans.  1-|-|. 

4.  Find  the  sura  of -5%,  |||,  }^,  and  ■^.      Ans.  2^. 
6.  Find  the  sum  of -^4,  -jW^,  and  ^^^.    Ans.  1^^. 

6.  Find  the  sum  of  i,  f ,  and  \.  Ans.  1^. 

7.  Find  the  sum  of  f ,  |,  and  ■^.  Ans.  1\^\. 

8.  Find  the  sum  of  |,  |f ,  and  |.  Ans.  2|4. 

9.  Find  the  sum  of  |,  ^,  and  ^^.  -4w5.  Iff. 

10.  Find  the  sum  of  \,  f ,  f ,  |,  and  |.  ^W5.  3-|. 

11.  Find  the  sum  of  7^,  8J,  and  3.  Ans, 

12.  Find  the  sum  of  16|,  12|,  and  ^^.  Ans. 


ADDITION".  141 

13.  Find  the  sum  of  5|,  9^^,  and  llf  Ans.  26^, 

14.  Find  the  sum  of  7^,  3^,  and  21|.  A71S.  32^. 

15.  Find  the  sum  of  5-gig^,  7^,  and  9^^^^.  Ans. 

16.  Find  the  sum  of  21f ,  18|,  4  and  26f       Ans.  70fJ. 

17.  Find  the  sum  of  17i  10,  14|,  and  13|.      Ans. 

18.  Find  the  sum  off,  3jf  10|,  and  ^.        Ans.  15^^. 

19.  What  is  the  sum  of  1246f  pounds,  9464^^  pounds, 
and  204^2^  pounds  ?  Ans. 

20.  What  is  the  sum  of  1476  bushels,  46-^  bushels,  and 
978-1  bushels  ?  Ans. 

21.  Bought  4  pieces  of  muslin  containing  90-J-,  40J^,  33^, 
and  61 J  yards  respectively.     How  many  yards  in  all  ? 

Ans.  225 J  yards. 

22.  Four  men  weighed  respectively  150J,  101,  185J,  and 
114|  pounds.     What  was  their  entire  weight.      A7is. 

23.  My  farm  is  divided  into  .5  fields :  the  first  contains 
lOJ  acres,  the  second  12|  acres,  the  third  15|  acres,  the 
fourth  9^  acres,  and  the  fifth  52Jf  acres.  How  many 
acres  in  my  farm?  Ans.  100  acres. 

24.  A  man  has  3  sheep  ;  the  first  is  worth  |6f,  the  second 
t8|,  and  the  third  |9J.  How  many  dollars  are  they  all 
worth  ?  Ans. 

25.  John  is  11^  years  old;  Mary  is  2^  years  older  than 
John  ;  and  Henry  is  6-^  years  older  than  Mary.  How  old 
is  Henry  ?  Ans. 

26.  A  man  walked  25f  miles  on  Monday,  27|  miles  on 
Tuesday,  30^  miles  on  Wednesday,  33|  miles  on  Thursday. 
How  far  did  he  walk  in  all  ?  Ans.  117^|-  miles. 

27.  Jones  has  746-J^  acres  of  land,  Wilson  has  594-  acres 
more  than  Jones,  and  Williams  has  as  many  acres  as  both. 
How  many  have  all  ?  Ans.  3103^. 


SECTION    IV. 


S:  mm  T'R' ACT  rQ:IV 


183. 


0i*at^x:GitciXeX 


Example  1.  — From  ^  take  4. 

Solution.— Since  the  difference  between  3  units  and  1  unit  is  3 
units,  the  difference  between  3  sevenths  and  1  seventh  must  be  3 

JPJJOBX^JIfS. 

1.  From  A  take  ,s^;  A;  A;  A;  A;  A- 

2.  From  «  take  ^V;  A;  A;  A;  A;  A- 

Example  2. — From  |  take  }. 

Solution. — By  reducing  |  to  8ths  we  have  | ;  then  from  |  we  take 
I,  which  leaves  as  many  Sths  as  there  are  units  in  the  difference  be- 
tween 7  and  6,  which  is  1  unit.  Therefore,  if  from  |  we  take  |,  we 
have  left  |. 

3.  From  i|  take  |;  |;  i;  i;  J;  f 

4.  From  fj-  take  A;  ^s^;  i;  |;  J;  };  f 

Example  3. — From  |  take  f. 

Solution. — The  least  common  multiple  of  8  and  7  is  56 ;  f  =  ||, 
and  f  =  If.  35  fifty-sixths  less  24  fifty-sixths  is  11  fifty-sixths.  There- 
fore, if  from  f  we  take  f,  the  remainder  will  be  |^. 

5.  From  f  take  |;  i;  |;  f;  ^;  A- 

6.  From  A  take  i;  |;  f ;  -f^',  A;  A- 
Example  4.— From  10  take  3|. 

Solution.—  10  is  equal  to  9  +  8  eighths.    From  8  eighthi  take  5 

142 


SUBTRACTION. 


143 


eigJUhs,  and  we  have  3  eighths  left.    From  9  take  8  and  6  remains. 
Therefore  10  less  3|  =  6|. 

7.  How  much  is  11  -  2f  ?     12  —  3|  ?     13  -  7^  ? 

8.  How  much  is  15  —  3i  ?     20  —  4|  ?    24  —  5|  ? 
Example  5. — How  much  is  4J  —  2J  ? 

Solution.— 4|  ~  4||,  and  2|  =  2||.  Since  we  cannot  subtract  28 
thirty-dxths  from  27  thirty-sixths,  we  reduce  one  of  the  4  units  to 
thirty-sixths,  which  added  to  27  thirty-sixths  gires  63  thirty-sixths.  63 
thirty-sixtJis  less  28  thirty-sixths  are  35  thirty-sixths,  and  2  from  3  leaves 
1.    Therefore  4|  less  2|  =  1||. 

9.   How  much  is  4J  —  2^  ?    5^  —  2}  ?     6^  —  2^  ? 
10.   How  much  is  5J  —  3^  ?    ^  —  ^^    21-^  -  l^^^  ? 


^•IWi'itten  ^:^er  ci^feX-i 


Example  1. 


23 

47 


SOLUTION. 

17  _  23  —  17 

47""      47 


47 


What  is  if  less  H  ? 

Explanation. — Since  the  dif- 
ference between  23  units  and  17 
units  is  6  units,  the  difference  be- 
tween 23  47ths  and  17  47th8  must 
be  6  Ji.7th8,  or  :^,  Ana. 

Example  2. — How  much  is  ^  —  ^^  ? 

Explanation.— We  first 

reduce  ^^  and  ^^  to  SJ^ths 

and  find  ^^  =  51  .?-^<A«, 

51  SJfifhs.  and  x\  =  40  SJfiths. 

40  SJiOths.  Subtracting     from     51 

-,-,   a>,n^i  1 1       sloths,  40  54C^A«,  we  have 

Example  3.— Subtract  3^^  from  20. 

SOLUTION.  Explanation. — ^^  from  1  (|f), 

20       ^::;19_j_l:^19_|_i6       leaves  VV«   3  from  19  (amount  loft 

o  7    QiJi    Q_l_7,      *^^^^  reducing  1  to  16ths)  leaves 

^tV  —    ^  +  TT  —    ^  +  T8-      ig     Annexing  to  this  ^^,  we  have 


SOLUTION 
^   X   17  _  _51 

20  X  17  ~  340 
^  X  20  _  ^ 
17  X  20  ~"  340 


144 


FRACTIONS. 


Example  4. 

SOLUTION. 

^=   4+  i=  4  +  il 


From  16f  subtract  4j: 

Explanation.  —  Since 
the  fraction  in  the  minu- 
end is  greater  than  the  one 
in  the  subtrahend,  we  take 
1  from  16  and  add  to  the 
f,  making  Y".    Then  we 


Ans.  llfj- 


reduce  the  fractions  to  a  common  denominator,  subtracting  them  as  in 
Ex.  1 ;  then  subtract  the  integers,  and  prefix  the  latter  remainder  to 
the  former. 

Rule  1.— Reduce,  if  necessary ,  the  fractions  to  a  common 
denominator  ;  then  write  the  difference  of  the  numerators  over 
the  common  denominator, 

II. — If  mixed  numbers  are  to  he  subtracted,  subtract  the 
integers  and  fractions  separately ;  or  reduce  to  improper 
fractions  and  proceed  ly  Rule  I. 


rnoBL  EMS. 

From 

12.  fj  take    |.  Ans,  i^, 

13.  8|  take  2|.  Ans.   6^. 

14.  9f  take  5f.  Ans.  4^. 

15.  lOf  take  5}.  Ans.  5^. 

16.  7|  take  2|.  Ans.  4|. 

17.  11  take  6|.  Ans.  4|. 

18.  15i  take    -J.  Ans.  l^, 

19.  9    take  3f  Ans,  5f 

20.  13i  take    f .  Ans,  12if. 

21.  12^  take    7f .  Ans,  4^^. 

22.  16f  takelOfI-  Ans,5^. 


From 

1.  \^  take  ^, 

2.  a  take  ^. 

3.  i  take  f 

4.  i  take  f . 

5.  f  take  f. 

6.  -f  take  f 

7.  -J  take  f 

8.  t  take  |. 

9.  ^take  |. 

10.  ^  take  f 

11.  H  take  J. 

23.  If  a  ton  of  hay  costs  $24f  and  a  ton  of  coal  $59J, 
what  is  the  difference  in  their  cost  ?  Ans.  $35^. 

24.  A  farmer  having  300  acres  of  land,  sold  149|  acres. 
How  many  acres  had  he  left  ?  Ans.  150^  acres. 


Ans. 

f 

Ans. 

A- 

Ans. 

i- 

Ans. 

A- 

Ans. 

T^- 

Ans. 

A- 

Ans. 

1- 

Ans. 

|. 

Ans. 

U- 

Ans. 

A- 

Ans. 

A- 

SECTION    V. 


m  MULTIPLICATION  r^ 


183. 


Oral  TK\:oi  ri^oj^ 


-•♦•- 


Example  1. — If  1  pound  of  coffee  costs  J  of  a  dollar,  how 
much  will  3  pounds  cost  ?    5  pounds  ?     7  pounds  ? 

Solution. — If  1  pound  of  coffee  costs  ^  of  a  dollar,  3  pounds  wOl 
cost  3  times  as  much,  or  f  of  a  dollar  (161).  5  pounds  will  cost 
5  times  as  much  as  1  pound,  or  f  =  1^  dollars. 


PR OBLEM8  . 

1.  At  1^  per  pound,  how  much  will  2  pounds  of  sugar 
cost?    3  pounds?    4  pounds?    5  pounds?    8  pounds? 

2.  At  $f  per  gallon,  what  cost  3  gallons  of  molasses? 
5  gallons  ?     7  gallons  ?    9  gallons  ? 

Example  2.— If  1  barrel  of  flour  costs  $8,  what  will }  of 
a  barrel  cost?    J?    -V-?    i^?    iy^? 

Solution.— If  one  barrel  of  flour  costs  $8,  i  of  a  barrel  will  cost  \ 
of  $8,  or  $2 ;  and  f  of  a  barrel  will  cost  3  times  |3,  or  $6.  |  of  a  bar- 
rel will  cost  7  times  ^  of  $8,  or  $7  ;  &c.,  &c. 

3.  If  an  acre  of  land  is  worth  $2400,  what  is  \  of  it  worth  ? 
i?    4?    J?    i?    i?    ^^?    JV? 


10 


145 


146 


FRACTIONS 


4.  If  one  yard  of  silk  is  worth  $4.80,  what  is  J  of  a  yard 
worth?    i?    I?    1?    A?    A?    A? 

Example  3. — A  man  owning  f  of  a  mill,  sold  f  of  his 
share.     What  part  of  the  mill  did  he  sell  ? 

Solution. — If  a  man  owning  f  of  a  mill,  sells  §  of  his  share,  he  sells 
I  of  f  of  the  mill.  ^  of  f  of  the  mill  is  3^^  of  the  mill,  and  f  of  f  of 
the  mill  are  2  times  -j'^,  or  /^  of  the  mill.  Therefore,  he  sells  /^  of  the 
mill. 

5.  How  much  is  f  multiplied  by  f  ?    By  f  ?    By  |^  ? 

6.  ixf=?     |xi=:?    |xi  =  ?    ixi  =  ?    |x|  =  ? 


tlWi  itteri  ^Xjer  ei;*e^  t 


Example  1. 


5M  =  of, 


-Multiply  H  l>y  7. 

Explanation. — Since  multi- 
plying the  numerator  hy  a  whole 
number,    the    denominator    re- 
maining   the    same,   multiplies 
the  value  of  the  fraction  (161), 
we  multiply  17  by  7  and  write 
the  product  119  over  the  denom- 
inator, and  thus  have  y^  =  5^f 
=  5|.     Or. 
Since  dividing  the  denominator  by  a  whole  number,  the  numerator 
remaining  the  same,  multiplies  the  fraction  (161),  we  divide  the 
denominator  21  by  7,  and  write  the  result  under  the  numerator ;  thus. 


17  X 
21 

17 

21-4- 


SOLUnON 

7_n9_ 
—  21  ~ 
or 


17 

,  =  i  =  'i- 


Example  2.— Multiply  42  by  ^. 

SOLUTION. 

42  X  13  _  42  X  13  _  546 
17  ~  17  ~  17 
or 

42  X  13  _  42  X  13  _  546 
17 ""   17   ~"  17 


=  32-ft, 


=  32^. 


Explanation. — We 
are  required  to  find  the 
same  part  of  42  that  |f 
is  of  1. 

We  can  do  this  in 
two  ways  : 

1st.  We  can  find  ^ 


MULTIPLICATION.  147 

of  42  =  ff,  and  then  multiply  yV  of  42,  or  f  f ,  by  13  to  find  ^|  of  42, 
which  equal  32^. 

Or,  2d.  We  can  multiply  42  by  13  ;  but  13  is  seventeen  times  as 
great  as  our  multiplier  jf.     Therefore  to  obtain  the  true  product  we 

must  divide  42  x  13  by  17,  and  we  have    —-- —  =  32j\,  Ana. 

Example  3. — Multiply  -j^  by  |^. 

SOLUTION.  Explanation. — ^We  are  required  to  find 

5         11         55  ii  of  A-     Is*-  We  can  find  ^V  of  iif  ^7 

T^  ^  22  ^^  156*  multiplying  the  denominator  by  12  (162), 

jf ^.     Having  found  ^f ^  to  be  3^,  we  find 
\^  by  multiplying  y|^  by  11  (161),  obtain- 
ing for  an  answer,  ^f^.     Or, 

We  may  multiply  ^^^  by  11  (161),  giving  us  f|.  But  we  were  to 
have  multiplied  by  \\,  a  number  only  yV  as  large  as  11  ;  therefore,  we 
must  divide  our  product  by  12  (162).  This  will  give  the  correct 
result,  ^^^. 

Example  4.— Multiply  21J  by  5. 

solution. 
21f  X  5  =  21  X  5  +  }  X  5  =  105  +  3f  =  108J. 
Explanation. — We  multiply  21  by  5,  and  |  by  5,  and  add  the 
two  products. 

Example  5. — Multiply  4|  by  7|. 
solution. 

Explanation. — We  reduce  both  factors  to  improper  fractions,  and 
then  multiply  as  in  Example  3. 

17      7      119 
Since  Example  1  can  be  solved  thus,  oT  ^  1  ~  "2r '    ^^^™P^®  ^' 

tt„«,  I  X  ^  =  ^;  and  Example  4,  thu.,  31}  x  5  =  ^  x|=^; 

by  changing  the  integers  to  a  fractional  form  by  writing  1  for  the 
denominators,  we  can  readily  see  by  inspection,  that  Examples  1  to  5 
may  all  be  solved  by  the  following 


148  FKACTIOIfS. 

Rule. — Reduce  each  integer  and  mixed  number  to  the 
form  of  a  simple  fraction ;  then  write  the  product  of  the 
numerators  over  the  product  of  the  denominators. 

To  lessen  labor,  cancel  all  factors  common  to  both  terms,  before 
multiplying. 

The  following  special  rules  can  be  deduced,  and  may  be  used,  if 
thought  best : 

To  multiply  a  fraction  by  an  integer,  or  an  integer  by  a  fraction, 
we  deduce  from  Examples  1  and  3, 

Rule  I. —  Write  the  prodtLd  of  the  integer  and  the  numerator  of  the 
fraction  over  the  denominator  ;  or, 

Write  the  quotient  of  the  denominator  of  the  fraction  divided  by  the 
whole  number  under  the  numerator. 

To  multiply  a  fraction  by  a  fraction,  from  Example  3,  we  deduce. 
Rule  II. —  Write  the  product  of  the  numerators  over  the  proditet  of 
the  denominators. 

To  multiply  a  mixed  number  by  an  integer,  we  have,  from  Exam- 
ple 4, 

Rule  III. — Multiply  the  whole  number  and  the  fraction  separately 
by  the  whole  number ,  and  add  the  products. 

To  multiply  mixed  numbers,  from  Example  5,  we  have 
Rule  IV. — Reduce  mixed  numbers  to  improper  fractions  and  apply 
Rule  II. 

Reducing  Compound  to  Simple  Fractions  is  the  same  as 
multiplying  one  fraction  by  another  (158). 


PB  o  a  z  EM 


Multiply 

1.  ^  by  3.  Ans,  If 

3.  ^  by  4.  Ans.  1-^. 

3.  m  by  5.  Ans.  3^. 

4.  tt  by  6.  Ans.  l^V 

5.  ^  by  9.  Ans.  JJ. 


Multiply 

6.  -^  by  10.  Ans.  2H 

7.  m  by  7.  Ans.  6ff| 

8.  fHbyS.  Ans.'Hm 

9.  -^  by  9.  Ans.  1^ 
10.  4  by  20.  Ans.  K\i 


MULTIPLICATION. 


149 


Multiply 

Multiply 

11.  28  by  J. 

Ans.  21. 

17.  165  by  |. 

Ans.  103^. 

12.  315  by  |. 

Ans.  210. 

18.  283  by  f 

^7i5.  220^. 

13.  118  by  f 

Ans.  59. 

19.  292  by  A- 

^W5.  87f 

14.  430  by  f 

^ws.  258. 

20.  568  by  A- 

Ans.  154^. 

15.  348  by  f 

Ans.  290. 

21.  443  by  t^j^. 

^ws.  258t«^. 

16.  756  by  f 

^ws  432. 

22.  687  by  H- 

Ans.  377H. 

23.  H  by  J. 

Ans.  i<ft. 

26.  ^  by  A. 

^W5.    TVff- 

24.  if  by  f 

^ws.  T^^. 

27.  «i  by  U- 

^^5.  -AWr 

25.  tt  by  H. 

^W5.  if|. 

28.  «by,ft. 

^ws.  J. 

29.  I  X  I  X  f  =  ?  Jw5.  f 

30.  i  X  i  X  H  =  ?  ^7i5.  f 

31.  A  X  H  X  H  =  ?  Ans.  a. 

32.  H  X  H  X  M  =  ?  ^^^5.  A- 

33.  H  X  If  X  Hi  =  ?  Ans.  H- 

34.  |xixixix4=?  ^/J5.  f 

35.  -J  X  f  X  A  X  H  X  ii  =  ?  ^W5.  T^. 

36.  «  X  H  X  it  X  H  X  il  =  ?  Ans.  «. 

37.  ixAx|xA  =  ?  ^^«-  iSAr- 

38.  A  X  t  X  1|  X  1|  X  4  =  ?  Ans.  4. 

39.  -H  X  H  X  lA  X  lA  =  ?  ^^^-  1- 

40.  What  cost  ff  of  a  bushel  of  oats,  at  ^  per  bushel  ? 

41.  A  man  owning  -ff  of  an  oil  well,  sold  f  of  his  share. 
What  part  of  the  well  did  he  sell  ?  Ans.  if. 

42.  B's  city  lot  contained  m  of  an  acre,  and  he  sold  J  of 
it.    What  part  of  an  acre  did  he  sell  ?  Ans.  4||. 

43.  What  cost  7-^  pounds  of  beef,  at  16  cents  a  pound  ? 

Ans,  11.21. 

44.  8|  dozen  hoes,  at  $9  per  dozen?  Ans.  $78. 


1>>0  rRACTio]srs. 

45.  9|  tons  of  sugar,  at  $180  per  ton  ?         Ans.  $1728. 

46.  15|  yards  of  silk,  at  $5  per  yard  ? 

A71S.  $7GJ  =  $76.88+. 

47.  20|  bushels  rye,  at  $1.20  per  bushel? 

Ans.  $24|  =  $24.75. 


Multiply 

Ans. 

Multiply 

Ans. 

48.  9^  by  5|. 

54^. 

52.  10-^  hy  llf. 

121A. 

49.  18|  by  4|. 

80H. 

53.  12J  by  16f 

203^. 

60.  16|  by  3i. 

57H. 

54.  131J-  by  16f. 

2187i 

51.  17i  by  12*. 

230^. 

55.  62J  by  137i. 

8593f. 

66.  What  cost  19|  cords  of  wood,  at  $3|  per  cord  ? 

Ans.  $66.23  +  . 
57.  What  cost  93^  bushels  oats,  at  33J  cents  per  bushel? 

A71S. 

68.  What  will  8f  weeks  board  cost,  at  $6J^  per  week  ? 

Ans. 

69.  At  12^  cents  each,  what  cost  12^  pints  of  peanuts  ? 

Ans.  $1.56i-. 

60.  What  is  the  value  of  204^^  acres  of  land  at  $72}  an 
acre  ?  A  ns, 

61.  What  is  the  value  of  580-|-  pounds  of  sugar  at  9^^  cents 
per  pound  ?  Ans. 

62.  What  will  85^  pounds  of  tea  cost  at  $1.37^  per 
pound?  Ans.  $117.56^. 

63.  Bought  156  pounds  of  cheese  at  $.12^  per  pound, 
327  pounds  of  coffee  at  26f  cents  per  pound,  and  17  barrels 
of  apples  at  $2.87|^  per  bbl.  What  was  the  cost  of  the 
whole  ?  Ans.  $155.84|. 

64.  What  is  the  value  of  5|  pieces  of  cloth,  each  piece 
containing  36}  yards,  at  $.85  per  yard?        Ans. 


SECTION    VI 


^ 


©I^riSI^M  I 


184. 


Example  1. — If  I  divide  f  of  a  dollar  equally  among  4 
boys,  what  part  will  each  boy  receive  ? 

Solution.— Each  boy  will  receive  J  of  f  of  a  dollar,  or  f  of  a 
dollar  (162). 

Pn  OBZJEMS. 

1.  Divide  ff  by  2.    By  3.    By  4.    By  6.    By  9. 

2.  Divide  if  by  2.     By  8.    By  12.     By  16.    By  24. 

Example  2.— Divide  4  by  4.     By  5.     By  7.    By  9. 
Solution. — |  divided  by  4  are  -^  (162) ;  f  divided  by  5  are  ^;  &c. 

3.  Divide  fj  by  3.     By  4.     By  9.     By  6.     By  7. 

4.  Divide  f  by  5.    By  6.    By  7.    By  9.    By  3. 

Example  3. — Divide  5  by  J. 

1st  Solution. — Since  5  =  *^,  and  since  20  units  divided  by  3  units 
equal  ^,  20  fourths  divided  by  S  fourths  must  equal  ^,  or  6|  (176). 

2d  Solution. — If  we  divide  5  by  3  we  have  f ;  but  our  divisor  is 
only  ^  of  3  ;  therefore  to  obtain  the  correct  quotient  we  must  multiply 
f  by  4,  giving  us  ^  =  6|. 

161 


152  FKACTION^S. 

6.  Divide  8  by},    f     f    i^.    |.    ^.    ^V    h    A- 

6.  Divide  11  by  f    f.     f    i.    f     A-    t^.-    A-    A- 
Example  4. — Divide  |  by  ^, 

1st  Solution.— I  =  ||,  and  TT=lf-  Since  77  units  divided  by 
72  M7iif«  =  ^,  77  eighty-eighths  divided  by  72  eighty-eighths  must  equal 
il,  or  1^. 

2d  Solution.— Dividing  i'bjQ  gives  ^V  (162) ;  but  our  divisor  is 
o^7  irr  of  nine,  therefore,  we  mast  multiply  if\  by  11  to  obtain  the 
true  result  which  is  ^  =  1/^. 

7.  Divide  A^yf   |.   h   i'   f   f    f   I   A-   A- 

8.  Divide  I  by  i.    f.    -J.    f    t^.    A-    f    h    f    A- 
Example  5.— Divide  4|  by  4.    5.     6.     7.    8.     9.    10. 

Solution. — 4  divided  by  4  is  1 ;  |  divided  by  4  is  ^.  Therefore 
4|  divided  by  4  equal  1/^. 

4|  divided  by  5  equal  ^  -j-  5  =  ff  ;  &c.,    &c. 

9.  Divide  7|  by  6.     7.    10.     12.    13.     15.    16.    8.    9. 

10.  Divide  9^  by  2.    3.     4.    5.     6.     7.    8.    9.    11.     12. 

Example  6.— Divide  5|  by  6f.     7J.    8^. 

Solution.— 5f  =  ^  =  f| ;  6|  =  ^  =  ff .  69  twelfths  divided  by 
80  twelfths  =  If. 

5|  =  ^ ;  7i  =  ^ ;  and  -^^  divided  by  »/  =  || ;  &c. 

11.  Divide  9|  by  3f    2|.    3|-.     5|.    4^-     H-     6^. 

12.  Divide  4^  by  2^.    3^     6J.    5f     9f    3^.    8^. 


Example  1.— Divide  ^  by  7. 

l8T  solution.  Explanation. -We  may 

f2:  "^  '  _£_  either  divide  the  numerator 

31  ~  31  21  by  7  (162) ;  or. 


DIVISION, 


153 


21 
31 


31   • 


2d  solution. 

Multiply  the  denominator 

21         3 

by  7  (162);  or. 

X  7  ~  217  ~  31. 

We  may  reduce  the  divisor 

3d  solution. 

to  217  31st8,  and  then  we  have 

21       217        21 

3 

21  Slsts  divided  by  217  SUts 

31    •    31   ~  217  " 

"31. 

=  iii  =  -h- 

Example  2.— Divide  22  by  f . 


1st  solution. 


22 


22 


99. 


=  99. 


Explanation— In  the  first  Solution, 
we  first  divide  22  by  2,  which  gives  us 
11 ;  but  in  dividing  by  2,  we  use  a  di- 
visor 9  times  too  large,  and  our  quo- 
tient is  only  \  of  what  it  should  be. 
We  therefore  multiply  it  by  9  and  ob- 
tain the  correct  result,  99. 

In  the  second  Solution,  we  multiply 
22  by  9  and  divide  the  product  by  2. 
This  gives  the  same  result  as  the  first 
Solution,  and  is  more  convenient  when 
the  integer  is  not  exactly  divisible  by  the  numerator. 

In  the  third  process,  we  reduce  both  dividend  and  divisor  to  the 
same  denominator,  9.    Then  198  ninths  divided  by  2  ninths  =  ^^^ 


a     23    - 

2d  solution. 
2  _  22  X  9 
9~      2 
3d  solution 


22 -^ 


198 
9 


=  99. 


5^ 
16 


1st  solution. 

5       5 

X 


4  = 


5         r, 


_5^ 
16 


64'  64 
2d  solution. 
_35     35 
"~16'  16 
3d  solution. 
4_^^^ 
7~112    •   112 


7  = 


35 

64. 


Example  3.— Divide  ^  by  4- 

Explanation.— Since  we   are 
35     required  to  divide  y^  by  4  sevenths^ 
g^    in  the  1st  Solution,  we  divide  ^^ 
by  4,  which  gives   us  ^  (162). 
But  as  4  is  seven  times  as  large  as 
4,  ^  is  only  one  seventh  of  the  re- 
quired answer.    Therefore  multi- 
plying g\  by  7  will  give  the  correct 
35     result,  which  is  ff. 
64,        In  the  2d  Solution,  we  first  mul- 
tiply by  7  and  then  divide  by  4, 
obtaining,  of  course,  the  same  result. 

In  the  3d  Solution,  we  reduce  -^-^  and  |  to  ^y'^  and  ^  respectively. 
Since  35  units  divided  by  64  units  gives  f|,  -^  divided  by  ^  must 
give  f  f . 


154  FEACTIONS. 

Example  4. — Divide  41J  by  9. 

1st  solution.  Explanation.— This  Ex- 

167  167         .,     ample  is  performed  as  Ex- 

41f  -T-  9  =  —  -J-  9  =  —  =  4|f .    ample   1,  after  the  mixed 

o^  o^TTTmx^,.T  number  has  been  reduced 
2d  solution. 

9  ^  41 3  ^  ^  fraction. 

l IL  It  is   sometimes  prefer- 

4|-|  able,  however,  to  solve  it 
thus :  9  is  contained  in  41  f 

4  times, with  the  remainder  5f  =  -\'-.  9  is  contained  in  ^,  |f  of  a  time. 
Therefore,41|  divided  by  9  -  4f f. 

Example  5.— Divide  18|  by  7|. 

solution. 

65 

18^  -;-  7f  _    ^      .    ^  _    ^    X  ^^  _  ^  _  2Aj. 

19 

Explanation. — We  first  reduce  the  mixed  numbers  to  simple 
fractions,  and  then  proceed  as  in  Example  3, 1st  Solution.  Of  course, 
by  the  2d  or  3d  Solution  the  result  would  be  the  same. 

The  Solution  of  Examples  1,  2,  and  4  may,  by  supplying  a  denomi- 

21       7 
nator  1  to  the  integers  in  each,  be  efiected  thus  :   Ex.  1.  ^  -*-:§■  = 

ol       X 

21     1       3     „„    22      2      22     9      „„„..,,       9      167 
_x^  =  gj    Ex.2.  ^^g=j-x-  =  99.    E^4.   41}^j  =  -3-x 

-  =  -^  =  4|f .    By  an  inspection  of  these  three  Solutions  and  the  let 

Solutions  of  Examples  3  and  5,  we  see  that  in  every  instance,  after 
changing  the  integers  and  mixed  numbers  to  fractions,  we  obtain  our 
answer  by  multiplying  the  dividend  by  the  divisor  with  its  terms  ivr 
verted,  that  is,  with  the  numerator  and  denominator  irUerchanged. 

The  Reciprocal  of  a  number  is  1  divided  by  that  number,  so  that 
the  Reciprocal  of  a  fraction  is  'the  same  as  a  fraction  with  its  terms 
inverted ;  thus,  f  is  the  reciprocal  of  |.     \  is  the  reciprocal  of  f  or  2 ; 

|off;  Aof¥;&c. 

Hence,  in  all  cases  of  division,  in  which  dividend,  or  divisor,  or 
both,  is  a  fraction,  we  can  solve  the  problems  by  the  following 

Rule. — Reduce  integers  and  mixed  numbers  to  the  form 
vf  simple  fractions  ;  and  then  multiply  the  dividend  hy  the 


DIVISION. 


155 


divisor  with  its  terms  inverted ;  or,  by  the  reciprocal  of  the 
divisor. 

We  may  also  have  the  following  Special  Rules :  From  Ex.  1,  1st, 
2d,  and  3d  Solutions,  we  deduce  for  dividing  a  fraction  by  a  whole 
number,  this 

Rule  I. — Divide  the  numerator  by  the  whole  number,  and  vyrite  the 
quotient  over  the  denominator.    Or, 

Multiply  the  denominator  by  the  whole  number,  and  write  the  product 
under  the  numerator.    Or, 

Reduce  the  fraction  and  whole  number  to  a  common  denominator., 
and  write  the  numerator  of  the  latter  under  that  of  the  former. 

To  divide  a  whole  number  by  a  fraction,  we  deduce  from  Ex.  2, 1st., 
2d,  and  3d  Solutions,  the  following 

Rule  II. — Divide  the  whole  number  by  the  numerator,  and  multiply 
the  quotient  by  the  denominator.    Or, 

Multiply  the  whole  number  by  the  denominator,  and  divide  the  product 
by  the  numerator.     Or, 

Reduce  the  whole  number  and  the  fraction  to  a  common  denominator, 
and  divide  the  numerator  of  the  former  by  that  of  the  latter. 

To  divide  a  fraction  by  a  fraction,  we  deduce  from  Ex.  3, 1st,  2d, 
and  3d  Solutions, 

Rule  III. — Divide  the  dividend  by  the  numerator  of  the  divisor,  and 
multiply  the  quotient  by  the  denominator  of  the  divisor.    Or, 

Multiply  the  dividend  by  the  denominator  of  the  divisor,  and  divide 
the  product  by  the  numerator  of  the  divisor.    Or, 

Reduce  both  fractions  to  a  common  denominator,  and  then  divide  the 
numerator  of  the  dividend  by  the  numerator  of  the  divisor. 

Reducing  Complex  to  Simple  Frtictions  is  the  same  as  dU 
f>iding  one  fraction  by  another  (157). 


PROBLEM, 


Divide 

2.  14^  by  20. 

3.  14  by  54. 

4.  i|4  by  108. 

5.  «  by  72. 

6.  ^  by  13. 


-AnS.    yfy 

Ans.  ^, 
Ans,  yIy 
Ans.  ^, 
Ans.  yi^. 
Ans.  ^, 


Divide 

7.  m  by    96. 

8.  1^  by  203. 

9.  ill  by  213. 

10.  -^t  by    24. 

11.  'm  by  144. 

12.  i,V/  by    91. 


156 

FRACTIONS. 

13. 

16  by  f 

Ans,    20. 

18.  483  by  ^. 

^m  536|. 

14. 

24  by  f . 

Ans,    27. 

19.  160  by  ^. 

^W5.  202f. 

15. 

36  by  A. 

Ans.    66. 

20.  272  by  |f 

Ans.  335^?.. 

16. 

401  by  f . 

^n5.  467f 

21.  256  by  H. 

Jw5.  304if. 

17. 

453  by  f|. 

^715.  578f. 

22.  365  by  |f 

^ws.  414^. 

23.  At  $^5jj.  per  pound,  how 
be  bought  for  $25  ? 

24.  If  one  gallon  of  molasses 
can  be  purchased  for  $150  ? 

25.  At  $Jf  apiece,  how  many 
for  $39  ? 


many  pounds  of  cofiee  can 

Ans.  100  pounds, 
cost  $3^^,  how  many  gallons 

Ans.  480. 
chickens  can  be  purchased 
93. 


Divide 

Divide 

26.     5|    by    8. 

Ans.  yV 

31.    24I3-V    by 

21. 

27.     5i    by    8. 

Ans.   jr. 

32.  7462J      by 

27. 

28.  10|    by    9. 

Ans.  H- 

33.     372JV    by 

27. 

29.  72VV  by    6. 

Ans.  12:rb. 

34.  146741    by 

140. 

30.  143  A  by  11. 

Ans.  13^. 

35.  3146^1    by 

324. 

36.     8f  by    24. 

Ans.  3i. 

40.    8i    by    3f . 

Ans. 

2A. 

37.  15|  by    8^. 

Ans.  If. 

41.19i    by    11 

Ans. 

lOf. 

38.  10|  by    9f. 

Ans.  If 

42.73i    by    9i. 

Ans. 

n- 

39.  12J  by  16|. 

Ans.    }. 

43.  54JI  by  25f 

Ans. 

H. 

44.  If  7f  yards  of  ribbon  cost  $4^|,  what  is  the  price  per 
yard?  Ans.  $f 

45.  If  2}  yards  of  cloth  cost  $17^^,  what  is  the  price  per 
yard?  Ans.  $6f 

46.  If  3|  gallons  of  oil  cost  $2^^^,  what  is  the  price  per 
gallon?  A71S.  ^. 

47.  How  many  yards  of  carpet  2|  ft.  wide  will  be  required 
to  cover  a  floor  14^  feet  in  length  and  lOj^  feet  in  width  ? 

Ans.  19^. 


DIVISION". 

Divide 

Divide 

48.    1   by   i. 

Ans.  2^. 

53.     li    by    a- 

49.    i   by    I 

^7i5.      f 

54-     it    by  -Wr- 

50.  A  by  f. 

^ws.  1}. 

55.    IH    by    If. 

51.  A  by  H. 

^/is.    |. 

56.    m   by    H 

52.  H  by  if. 

Ans.  7^. 

57.  TftWrbyiftWr 

157 


58.  Find  the  value  of(f-J)-^(|-  —  f).       ^?^s.  If 

59.  Find  the  value  of  (i -^  %)  +  (i -^ -f^).  Ans.  2fJ. 

60.  Find  the  value  of  (|  x  i)  -r-  ( J  x  |).         Ans.  f 

61.  Find  the  value  of  (^  x  i)  x  (f  x  M).  Ans.  ^V 

62.  Find  the  value  of  |,  or  of  ^  -^  f .  Ans.  fj. 

63.  Find  the  value  of  |  of  |,  or  of  f  x  |.  Ans,  i. 


64.  Find  the  value  of 


4  +  1' 


orof  (|-|)~(J+|). 


65.  Find  the  value  of  t^^>  or  of  (|  x  H)  "^  (i  x  i)- 


Ans,  11. 


ItEVIEW    PROBIjEMS, 


Oi'^it  5^Xoi'cis^o^ 


1.  Jones  owns  J  of  a  rolling  mill,  Markham  f  and  Simp- 
son ^ ;  how  much  do  they  all  own  ? 

2.  Smith  sold  -^  of  his  interest  in  his  business  to  one 
of  his  clerks,  and  ^  to  another.  How  much  did  he  sell  to 
both? 

3.  After  paying  out  \  of  my  money,  and  depositing  f  of 
it  in  bank,  what  part  had  I  left  ? 

4.  Burnside  sold  at  one  time  f  of  his  farm,  at  another 


168  FBACTIONS. 

■^^,  at  another  time  -^;   how  much  of  his  farm  had  he 
remaining  ? 

5.  Hamilton  owned  f  of  a  planing  mill ;  he  sold  f  of  his 
interest  to  Boyd.     What  part  of  the  mill  did  he  sell  Boyd  ? 

6.  James  lost  -f  of  his  marbles.  He  then  had  28.  How 
many  had  he  at  first  ? 

7.  A  man  purchased  a  steamboat,  paying  at  the  rate  of 
$600  for  ^  of  ^  of  it.    How  much  did  the  boat  cost  him  ? 

8.  A  w^oman  purchased  21  yards  of  silk  for  $75.  What 
did  she  pay  per  yard  ? 

9.  James  bought  9  oranges  for  58  cents.  What  did  he  pay 
apiece  for  them  ? 

10.  A  grocer  sold  20^  pounds  of  lard  for  $4.10.  What 
was  that  per  pound  ? 

11.  At  35^4  cents  per  pound,  how  many  pounds  of  copperas 
can  be  bought  for  66  cents  ? 

12.  After  selling  3^  acres  to  A,  and  5-|  acres  to  B,  and  4J 
acres  to  0, 1  find  that  I  have  |  of  my  land  left.  How  much 
had  I  at  first  ? 

13.  One  bushel  is  the  ^^  of  ^  of  my  entire  crop  of  oats. 
How  many  bushels  of  oats  did  I  raise  ? 

14.  A  suit  of  clothes  cost  $50,  and  the  price  of  a  pair  of 
boots  was  ^  of  the  cost  of  the  suit.  What  was  the  price  of 
the  boots  ? 

15.  If  7  men  eat  6  loaves  of  bread  in  1  day,  how  many 
loaves  will  1  man  eat  in  the  same  time  ?    2  men  ?     10  men  ? 

16.  If^oi^  of  the  price  of  my  farm  equals  the  value  of 
my  dwelling,  and  my  dwelling  is  worth  $375,  what  is  the 
price  of  my  farm  ? 

17.  If  7  yards  of  broadcloth  cost  $40 J,  how  much  does 
1  yard  cost  ? 

18.  What  cost  7i  pounds  of  sugar  at  12  J  cents  per  pound  ? 


DIVISIOK.  159 

19.  If  95f  cents  be  paid  for  12J^  pounds  of  rice,  what  cost 
1  pound? 

20.  If  A  can  do  J  of  a  piece  of  work  in  1  day,  and  B  can 
do  I  of  the  same  work  in  1  day,  what  part  of  it  will  A  and 
B  together  do  in  1  day  ?  In  how  many  days  will  they  per- 
form the  whole  work  ? 

21.  If  a  man  can  do  a  piece  of  work  in  5  days,  what  part 
will  he  do  in  1  day  ?    In  2  days  ?    In  3  days  ?    In  4  days  ? 

22.  A  can  build  a  wall  in  10  days,  and  B  can  build  the 
same  wall  in  12  days.  In  what  time  will  A  and  B  together 
do  the  work  ? 

23.  If  C  and  D  together  can  mow  a  field  in  3J  days,  and 
D  alone  can  mow  the  same  field  in  5^  days,  in  what  time 
can  0  alone  mow  the  field  ? 

24.  If  I  of  I  of  a  barrel  of  flour  cost  $1  J,  what  cost  1  bar- 
rel ?     H  barrels  ? 

25.  Paid  $3^  for  a  City  Directory,  and  2^  times  as  much 
for  a  Webster's  Unabridged  Dictionary ;  what  did  I  pay  for 
both? 

26.  A  market  woman  bought  eggs  at  3  for  5  cents,  and 
sold  them  at  4  for  7  cents.  How  much  did  she  gain  on 
each  egg? 

27.  A  confectioner  bought  oranges  at  30  cents  for  12, 
and  sold  them  again  at  the  rate  of  8  for  21  cents ;  and  then 
found  that  he  had  gained  64  cents.  How  many  oranges  did 
he  buy  ? 

28.  After  selling  200  bananas  at  a  loss  of  6j  cents  each,  I 
found  that  I  could  purchase  lemons  at  3  for  7  cents  and  sell 
them  at  7J  cents  apiece.  How  many  lemons  must  I  buy 
and  sell  to  make  good  my  loss  on  the  bananas  ? 

29.  How  many  pencils  at  7^.  must  be  paid  for  150  pens, 
at  the  rate  of  1 J  cents  apiece  ? 


160  FRACTIONS. 


t'WVl:*ittei\^?&r-ci^GXt 


1.  A  bought  a  farm  for  $3250,  and  sold  ^  of  it  to  B,  and 
the  remainder  to  C.    What  is  the  value  of  C^s  property  ? 

Ans.  $2375. 

2.  A  merchant  bought  25  pieces  of  muslin  at  $3:^  per 
piece  ;  125  pieces  prints  at  $2| ;  35  pieces  ticking  at  $7J. 
What  was  the  amount  of  the  bill  ?  Ans.  $716.25. 

3.  A  man  owns  a  farm  of  125^  acres,  worth  $62f  per 
acre;  another  of  83}  acres,  worth  $76^  per  acre.  How 
much  are  both  farms  worth  ?  Ans.  $14240.144-. 

4.  A  grocer  bought  5  barrels  of  granulated  sugar;  the 
first  barrel  containing  225 J  pounds,  the  second  221J^  pounds, 
the  third,  23 1|^  pounds,  the  fourth,  229-J-  pounds,  and  the 
fifth,  240  pounds,  at  11 J  cents  per  pound.  He  sold  }  of  it 
at  12i  cents  per  pound,  and  the  remainder  at  12}  cents  per 
pound.     How  much  did  he  gain  ?  Ans.  ^16.06 -{-. 

5.  From  3}  acres,  I  sell  to  A  f  of  an  acre,  to  B,  f  of  an 
acre,  to  0,  |  of  an  acre,  to  D,  f  of  an  acre.  How  much  is 
the  remainder  worth,  at  $2575-J  per  acre  ?     Ans.  $2042^^. 

6.  Into  how  many  house  lots,  each  containing  rf^  of  an 
acre,  can  I  divide  184  acres,  and  for  how  much  apiece  must 
I  sell  lots,  to  realize  $95600?   Ans.  150  lots;  price,  $637^. 

7.  A  stove  manufacturer  purchased  old  metal  at  $^^ 
per  hundred  pounds,  and  makes  out  of  it  stoves  weighing 
350  pounds  each,  which  he  sells  at  $17^^  apiece.  How 
much  does  he  gain  on  each  100  pounds  of  old  metal  ? 

Ans.  $4.13. 

8.  A  man  bought  a  farm  for  $17500,  and  sold  |  of  it  to 
Mr.  B  for  $4500 ;  \  of  the  remainder  to  Mr.  0  for  $4600; 


REVIEW     PROBLEMS.  161 

and  the  remainder  to  Mr.  L  at  a  profit  of  $1250.     What 
was  his  whole  gain  ?  A  ns.  $5975. 

9.  A  huckster  bought  125  bushels  of  apples  at  $J  per 
bushel ;  35  bushels  at  $J  per  bushel ;  and  25  bushels  at  ^ 
per  bushel.  At  what  price  per  bushel  must  he  sell  them  to 
gain$30.61i?  Ans.t.^Z 

10.  A  hunter  in  pursuit  of  a  fox  that  was  ^\  miles  ahead 
of  him,  ran  1^  hours,  at  the  rate  of  5  miles  an  hour.  The  fox 
advanced  at  the  rate  of  3f  miles  per  hour,  for  1^  hours.  How 
far  in  advance  of  the  hunter  is  the  fox  ?     Ans,  l^f  miles. 

11.  Henry,  William,  and  Jacob  can  remove  a  pile  of  wood 
in  4  hours ;  and  Henry  and  William  together  can  remove  it 
in  9|^  hours.  In  how  many  hours  can  Jacob  alone  remove 
it  ?  Ans.  Q\\  hours. 

12.  Two  men  115  miles  apart  start  toward  each  other,  and 
travel,  one  at  the  rate  of  2||-  miles  per  hour,  and  the  other 
at  the  rate  of  3J  miles  per  hour.  In  how  many  hours  will 
they  meet  ?  Ans.  18|4  hours. 

13.  L,  M,  and  N  bought  a  drove  of  cattle :  L  paid  for  f 
of  the  drove,  M,  for  -^  of  it,  and  N,  for  the  remainder.  It 
was  found  that  L  paid  $640  more  than  N.  What  did  each 
pay,  and  what  was  the  cost  of  the  drove  ? 

Ans.  L,  $3840 ;  M  and  N,  each  13200 ;  $10240. 

14.  How  many  barrels  of  apples  at  $9  per  barrel  will  pay 
for  95  cords  of  wood  at  $5J  per  cord  ?  Ans.  55^^. 

15.  A  gentleman  purchased  a  farm  for  $9000;  he  paid 
$95  for  fencing  and  other  improvements,  and  then  sold  it 
for  $381^  less  than  he  had  expended.  What  did  he  receive 
for  it  ?  Ans.  $8713.75. 

'\Q.  If  a  family  of  5  persons  consume  f  of  a  barrel  of  flour 
in  one  month,  in  how  many  months  will  they  consume  9.1 
barrels .?  Ans.  IZ^. 


162  fractio:n^s. 

17.  If  a  man  earns  $112  per  month  and  spends  177  in 
the  same  time,  how  long  will  it  take  him  to  save  $941  ? 

Ans.  26-|^  months. 

18.  There  are  two  outlets  to  a  cistern ;  by  one  15  gdftlons, 
and  by  the  other  20  gallons,  can  be  emptied  in  one  minute. 
How  many  gallons  may  be  emptied  by  both  in  17f  min- 
utes? 

19.  The  product  of  three  numbers  is  79.  The  first  num- 
ber is  7,  the  second  is  9|.    What  is  the  third  ?     Ans.  l^J^. 

185.  Converse  deductions. 

We  have  already  learned  that  Fractions  with  reference  to  the  mode 
ofexpremng  them  are  either  called  ** Decimals"  or  ** Fractions.'* 

In  the  decimal  form,  the  denomination,  or  fractional  unit,  is 
indicated  by  the  position  of  the  decimal  point  (119)  ;  but,  in  the 
fractional  forin,  the  denomination,  or  fractional  unit,  is  ex- 
pressed by  the  denominator'  (152). 

We  shall  now  proceed  to  learn  how  to  change  Decimals  to  Frac- 
tions and  Fractions  to  Decimals  under  the  heading  Co^iverse  He- 
duction. 

Example  1. — Reduce  .875  to  a  common  fraction. 

SOLUTION.  Explanation. — .875  is  read  875  thousandfJis. 

375         7     Therefore  our  72wwer«^c»r  is  875,  and  our  (fenow- 
.875  =  TT^QQ  =  Q     inator,  1000,  and  the  fraction,  yH^,  which  re- 
duced to  its  lowest  terms  by  Rule  (178)  =|, 
required  fractional  form. 

Example  2.— Reduce  .03-^  to  a  fraction. 

SOLUTION.  Explanation.  —  We 

31  10  10  1        write  the  number  with- 

.03^  =^^—  "3"  -i-  10^  =  goQ  =  3Q       out   the   decimal    point, 

and  express  its  fractional 
unit  by  the  known  denominator  100,  which  gives  us  the  complex  frac- 
tion ^~,  which  reduced  to  a  simple  fraction  by  Rule  for  Division  of 
lUU 

Fractions  =  ^,  required  fractional  fonn. 


CONVERSE     REDUCTIONS.  163 

Hence,  to  reduce  a  decimal  to  a  fraction,  we  have  the 
following 

EuLE. —  Write  the  given  numler  of  decimal  units.  Omit 
the  decimal  point,  and  express  the  fractional  unit  hy  a  de- 
nominator. Reduce  the  resulting  fraction  to  its  simplest 
form. 

Example  3.— Keduce  |  to  a  decimal. 

SOLUTION.  1st  Explanation. — Annexing  three  ciphers  to  7, 

7  8  )  7.000     reduces  it  to  thousandths,  and  |  of  7000  thousandths 

8  ^^  875      ~  ^'^^  thousandths  or  .875. 

2d  Explanation. — Since  f  expresses  the  quotient 
of  7  divided  by  8,  we  annex  decimal  ciphers  to  7,  and  divide  by  8,  as 
in  Division  of  Decimals. 

Hence,  to  reduce  a  fraction  to  a  dechnal,  we  have  the  fol- 
lowing 

KuLE. — Annex  a  decimal  cipher  or  ciphers  to  the  numera- 
tor and  divide  by  the  denominator. 

PROBLEMS, 


1.  Keduce  .625  to  a  fraction. 

3.  Reduce  .75  to  a  fraction. 

5.  What  fraction  =  .4375  ? 

7.  What  fraction  =  .008  ? 

9.  Reduce  .28125  to  a  fraction. 
11.  Reduce  .0375  to  a  fraction. 
13.  What  fraction  =  .56  ? 
15.  What  fraction  =  .096  ? 


2.  Reduce  |  to  a  decimal. 
4.  Reduce  f  to  a  decimal. 
6.  What  decimal  =  T^g.  ? 

8.  What  decimal  =  T^  ? 
10.  Reduce  -^  to  a  decimal. 
12.  Reduce  ^%  to  a  decimal. 
14.  What  decimal  =  ^  ? 
16.  What  decimal  =  ■^? 


17.  Reduce  ^  to  a  decimal.       Anf^.  .3-J^or.33-J-or  .333^  &c. 

Sometimes  the  decimal  is  interminable,  and  in  such  cases  a  fraction 
may  be  written  after  the  decimal  figures,  the  quotient  being  carried 
to  any  desired  number  of  decimal  places ;  or  the  sign  +  may  be 
written  after  the  decimal  figures,  to  show  that  the  divisor  is  not  exact. 

Decimal  figures  which  continually  repeat,  are  called  a  Repetend, 


164  FRACTIONS. 

The  value  of  a  repetend  is  expressed  by  a  fraction  whose  numer- 
ator is  the  repeating  figures  and  whose  denominator  is  as  many  nines 
as  there  are  figures  in  the  repetend.     Thus, 

i  =  .333  &c.  =  f II  =  i  ;  I  =  .66  &c.  =  ff  =  | ;  &c.,  &c. 

That  this  is  correct  is  shown  from  the  fact  that  \  produces  .1111 
&c.;  therefore,  f  will  produce  .222  &c. ;  |,  .33  &c.;  |,  .555  &c.  In  like 
manner  ^^  will  produce  .0101  &;c.;  therefore,  ^^  will  produce  .02  &c.,' 
&c.,  &c. 

18.  Eeduce  the  repetend  .384615  to  a  fraction. 

19.  Reduce  ^^  to  a  decimal. 

20.  Reduce  the  repetend  .238095  to  a  fraction. 

21.  Reduce  ^  to  a  decimal. 

22.  Reduce  the  repetend  .142857  to  a  fraction. 

23.  Reduce  ^  to  a  decimal. 

24.  Reduce  the  repetend  .428571  to  a  fraction. 

25.  Reduce  f  to  a  decimal. 

186.  Multiplication  and  Division  by  Ali^ 
quot  l^ai^ts. 

In  common  oral  transactions,  resort  is  often  had  to  short  methods 
of  calculations,  which  in  many,  though  not  by  any  means  in  all  cases, 
facilitate  calculations. 

We  shall  learn  these  short  processes  under  the  heading  Multijill' 
cation  and  Division  by  Aliquot  I*arts, 

An  Aliquot  l^art  of  a  number  is  any  exact  divisor  of  it.  Thus, 
6  is  i  of  12  ;  3  is  i  of  12  ;  &c. 

Tlie  Unit  of  an  Aliquot  Part  is  that  number  which  is  divided 
to  obtain  the  part. 

The  Aliquot  Parts  of  $1  are  as  follows  : 


5    cents  =  ^V  of  $1. 

16|  cents  =  i  of  $1.      ^ 

61      "      =  A   «    " 

20        "      =i  "    " 

The  Unit  of  the 

81      "     =tV   ''    " 

25        "      =i   •'    " 

>     Aliquot  Part 

LO       "      =  tV  "    " 

331-      .*     :^  ^  «    u 

is  1. 

3i      "     =i     "    " 

50       "     =i  "    "       J 

ALIQUOT     PABTS.  165 

MULTIPLICATION. 
Example  1. — What  cost  144  pencils,  at  $.06J  a  piece  ? 

Solution. — Since  6^  cents  =  -^^  of  a  dollar,  the  cost  of  the  pencils 
is  j\  as  many  dollars  as  there  are  pencils.  jV  of  144  is  9.  Therefore, 
l4i  pencils  cost  $9. 

PROBLEMS. 

1.  What  cost  32  yards  of  muslin,  at  1.1 2|^  per  yard? 

2.  What  cost  36.6  yards  sheeting,  at  16f  cents  per  yard? 

3.  What  cost  24  yards  of  alpaca,  at  33^  cents  per  yard  ? 

4.  What  cost  28  pounds  of  butter,  at  20  cents  per  pound  ? 

5.  What  cost  64  wooden  pails,  at  25  cents  per  pail  ? 

6.  What  cost  19  pounds  of  tobacco,  at  50  cents  a  pound? 

7.  What  cost  120  oranges,  at  10  cents  each  ? 

8.  What  cost  24  slates,  at  8-J-  cents  a  piece  ? 

9.  What  cost  60  gum  erasers,  at  5  cents  a  piece  ? 

10.  What  cost  128  bottles  of  ink,  at  6^  cents  a  bottle? 

In  like  manner  we  may  multiply  when  the  unit  of  the  aliquot  part 
is  100,  1000,  &c. 

Example  2.— What  cost  6 J  shares  of  mining  stock,  at 
$144  per  share  ? 

Solution.— At  $144  per  share  100  shares  cost  $14400,  but  since  6^ 
is  tV  of  100,  6^  shares  cost  ^V  of  $14400,  or  $900. 

11.  What  cost  12^  yards  of  alpaca,  at  32  cents  per  yard  ? 

12.  What  cost  16|  pounds  of  Java  coffee,  at  36.6  cents  per 
pound  ? 

13.  What  cost  33|  pounds  of  coffee,  at  24  cents  per  pound  ? 

14.  What  cost  20  pounds  of  butter,  at  28  cents  per  pound? 

15.  What  cost  25  wooden  pails,  at  64  cents  apiece  ? 

16.  What  cost  50  papers  of  tobacco,  at  19  cents  per  paper? 


166  FRACTIONS. 

17.  What  cost  10  pairs  of  gloves,  at  $1.20  a  pair? 

18.  What  cost  8-J-  pounds  spices,  at  24  cents  per  pound  ? 

19.  What  cost  5  quires  paper,  at  60  cents  a  quire  ? 

20.  What  cost  6^  gallons  berries,  at  11.20  per  gallon  ? 

DIVISIOIST. 

In  a  somewhat  similar  manner,  we  may  perform  certain  examples 
in  division. 

Example. — At  6J  cents  apiece,  how  many  pencils  can  be 

bought  for  $9  ? 

Solution. — Since  Q^  cents  =  ^^  of  $1, 16  pencils  could  be  bought 
for  $1 ;  and  for  $9, 9  times  16  pencils,  or  144  pencils,  could  be  bought. 

PM  OBL  EMS. 

1.  At  12-J^  cents  per  yard,  how  many  yards  of  muslin  can 
be  had  for  $4? 

2.  At  16|  cents  per  yard,  how  many  yards  of  sheeting  can 
be  purchased  for  $6.10  ? 

3.  At  33-J^  cents  per  yard,  how  many  yards  of  alpaca  may 
be  purchased  with  $8  ? 

4  At  20  cents  per  pound,  how  many  pounds  of  butter 
can  I  buy  with  $5.60? 

5.  For  $16,  how  many  wooden  pails  can  I  buy,  at  25 
cents  a  pail  ? 

6.  For  19.50,  how  many  pounds  of  tobacco,  at  60  cents 
per  pound,  can  I  purchase  ? 

7.  If  I  pay  $9  for  144  pencils,  how  much  is  that  apiece  ? 

8.  The  cost  of  32  yards  of  muslin  is  $4 ;  how  much  is  the 
cost  per  yard  ? 

9.  If  I  pay  $8  for  24  yards  of  alpaca,  how  much  is  that 
per  yard  ? 


BILLS. 


167 


187.    BILLS, 

A  Sill  is  a  detailed  written  statement  of  items,  their  prices,  &c., 
which  one  party,  called  the  Creditor,  renders  to  another  party,  called 
the  Dd)ior, 

The  party  purchasing,  or  owing,  is  the  Debtor ,  and  the  party 
from  whom  he  purchases,  or  whom  he  owes,  is  the  Creditor. 

When  the  biL  is  paid,  the  person  receiving  payment  writes  at  the 
bottom  of  the  bill  "  Rec'd  Payment,"  or  "  Paid,"  and  signs  the  name 
or  the  names  of  the  party,  or  parties,  for  whom  he  is  acting,  as  an 
acknowledgment  of  the  bill  having  been  paid.  The  bill  is  then  said 
to  be  receipted. 

EXERCISES. 

FIRST. 

Pittsburgh,  Nov.  21, 1876. 
Messrs.  B.  H.  Hlnchman  &  Sons, 

Bought  of  Pennsylvania  Lead  Co. 


40  Kegs  White  Lead  100s.  4000 

20  "    "    "    50s.  1000 

80  "    "    «    25s.  2000 

7000  lb.,@9|/. 

$691 

25 

Mr.  Wm.  Unger, 


SECOND. 

Pittsburgh,  April  1, 1876. 

Bought  of  W.  H.  McClintock  &  Co. 


54  yards  Ingrain  Carpet,  @,  $1.25 
25    "       Venetian  Border, "  $1. 
68    "       Carpet  Lining,     "12^^. 
Making    and    Laying  86  yards 

Ingrain  and  Border,  @  10!^. 
20  yards  Oil  Cloth,         @,  90<^. 
15     "      Binding,  "     5^. 

Rec'd  Payment,  ~ 

W.  H.  McClintock  &  Co., 

Per  James  Sater. 


Jan. 

28 

it 

tt 

29 

*t 

31 

$67 

50 

25 

8 

50 

8 

60 

18 

75 

$128 

35 


168 


FKACTIOKS. 


Calculating  by  aliquot  parts,  the  derk  will  carry  out  all  the  amounts 
of  the  Bill  mentally. 


Mr.  SmoN  Martin, 


THIRD. 

Pittsburgh,  April  1,  1876. 

To  Murphy  &  Hamilton,  Carpenters.  JDr. 


Feb. 

3 

« 

« 

Mar. 

5 

" 

" 

u 

31 

t( 

« 

120  feet  of  Flooring,  @  3/'. 

3  days'  time  of  Carpenter,  @  $3  50. 
1  day's  do.  do.  "  3.50. 
Repairing  Step  Ladder, 

250  feet  Flooring,  @,  3^^. 

4  days'  time  of  Carpenter,  @  $3.50. 

Rec'd  Payment, 


50 


$40 


85 


Mr.  Martin  Searle, 


Murphy  &  Hamilton. 

rOURTH. 

New  York,  May  3,  1876. 
To  New  York  Tribune,  Dr. 


Apr. 

9 

To  Adv.  25  lines,  3  times,  @  25f . 

$ 

<< 

10 

"      "       3  lines,  Local,  @  $1. 

K 

25 

"     "     14    '*      @37|^. 

<( 

<t 

«      "       5    "      Local,  @  75)^. 

$30 

75 

Prof.  John  Harper, 


FIFTH. 

Allegheny  City,  Jan.  11,  1876. 
To  Renters  &  Co.,  Dr. 


May 

13 

1  Spring  Bottom  Bed,  @  $35. 

$ 

« 

« 

6  Cane  Seat  Chairs,       "  $2. 

f( 

" 

2     "       "     Rockers,    "  $5. 

u 

« 

2  Walnut  Cottage  Bedsteads,  @  $9. 

$ 

BILLS, 


169 


Mr.  J.  R.  McCuKE, 


SIXTH. 

Philadelphia,  April  10, 1876. 

Bo't  of  W.  V.  MCCUTCHEON. 


Apr. 

9 

35  yards  Broadcloth,  @  $4.25, 
10      "     Silk,              "    1.50, 
13t    "     Flannel,       "      .75,     . 
50     "     Muslin,         **      .14, 

$ 

Paid,  April  13/76. 

R.  McCUNE, 

For  W.  V.  McCuTCHEON. 


SEVENTH. 


Mr.  R.  U.  Slater, 


Chicago,  June  1,  1876. 
Bo't  of  G.  Oglethorpe. 


135  pounds  Rio  Coffee,        @  2Qf. 

$33 

60 

350       "      Crushed  Sugar,  "  11^ 

37 

50 

67       "      Carolina  Rice,   "10/. 

6 

70 

37       "      French  Prunes, "  13|/. 

4 

63 

35       "      Y.H.  Tea,          "75/. 

18 

75 

$90 

08 

Rec'd  Payment, 

G.  Oglethorpe. 


EIGHTH. 


Messrs.  Hall  &  Kerr, 


Cleteland,  July  5,  1876. 
Bo't  of  Lyon,  Thorp  &  Co. 


16  Tons  Bar  Iron,     @  $106.13, 

8     "    Cast  Steel,  "    456.37, 

35  kegs  Nails,  **        9.50, 


$5586 


54 


170  FRACTIONS. 

9.  Make  out  a  bill  against  John  Wagner  for  the  following 
items,  purchased  of  James  Tyrone,  1876:  May  1,  2  setts 
harness,  @  $35 ;  2  saddles,  @  $15  ;  May  5,  3  wagon  hubs,  @ 
$1.50;  5  pairs  buckles,  @  $.75 ;  and  2  buggy  axles,  @  $8. 

Ans.  $124.25. 

10.  Make  out  bill  for  Joseph  Hoffman,  who  has  bought 
from  Edward  Stewart  in  New  York  the  following  items: 
1876,  Mar.  16,  13  yards  cashmere,  @  $1.25;  8  yards  serge, 
@  40^ ;  6  yards  muslin,  @  12^^ ;  11  do.  cambric,  @  10)^ ; 
April  8,  5  yards  prints,  @  lOj^ ;  6 J  do.  mushn,  @  12f ;  and 
6  do.  cambric  @  13/^.  Ans.  $23.36. 

11.  Mrs.  M.  Stewart  bought  of  S.  B.  Home  in  1876,  May 
11,  1  pair  kids,  $2.00;  2  sprays,  @  30)^;  1  yd.  Euching, 
@  50)^;  i  yd.  illusion,  @  80)^ ;  May  16,  2  umbrellas,  @  $2.50; 
4  yards  ribbon,  @  30f ;  4  pairs  hose,  @  75^ ;  4  yd.  elastic, 
@26f  ;  3  ounces  zephyr,  @  25^ ;  6  handk'fs,  @  40;^.  Make 
out  bill  and  amount.  Ans.  $ 

12.  A.  L.  Sturgeon,  Esq.,  bought  of  Samuel  Wallace  of 
St.  Louis,  1876,  May  9, 1  broadcloth  dress  coat,  $45  ;  1  vel- 
vet vest,  $10 ;  2  pairs  pants,  @  $14 ;  1  pair  suspenders,  $1.25 ; 
3  boxes  collars,  @  35^ ;  and  had  overcoat  repaired  at  a  cost 
of  $2.50.     What  was  his  bill  ?  Ans.  $87.80. 

13.  The  following  articles  were  sold  in  Pittsburg,  Pa.,  by 
A.  H.  Enghsh  &  Co.,  to  J.  R  Weldin  &  Co.,  March  28th, 
1876:  600  Progressive  Spellers,  at  $.15  each  ;  200  Progres- 
sive First  Readers,  at  $.16 J  each ;  100  Progressive  Second 
Readers,  at  $.35  each  ;  100  Progressive  Third  Readers,  at 
$.53^  each  ;  50  Dean's  Primary  Arithmetics,  at  $.16|  each  ; 
50  Dean's  Intellectual  Arithmetics,  at  $.36  each ;  25  Burtt's 
Practical  Grammars,  at  $.66|  each  ;  48  Lossing's  U.  S.  His- 
tories, at  $1.00  per  copy.    Make  out  a  receipted  bill. 


OUTLIl^E, 


171 


OUTLINE   OF   DENOMINATE   NUMBERS. 


m 
» 
» 


O 
Q 


DEFINITIONS. 


192.  MONEY. 


198.   WEIGHT. 


MEASURES. 


188.  SIMPLE  NUMBERS. 

1 89.  SIMPLE  DENOMINATE  NUM- 

BERS. 

190.  COMPOUND     DENOMINATE 

NUMBERS. 

191.  ^  MEASURE. 

193.  United  States. 

194.  Canada. 

195.  English. 

196.  Germnn. 
191.  French. 

199.  Troy. 

200.  Apothecaries. 

201.  Avoirdujtois. 

202.  Common. 

203.  Surveyors''. 

204.  Common. 

205.  An  Angle. 

206.  A  Right  Angle. 

207.  Rectangle. 

208.  Square. 


Linear. 


-I 


iSguare. 


Terms. 


Cubic. 


209.  Measure. 

210.  Aj-ea. 

211.  Surveyors''. 

212.  Common. 

(-213.  A  Cube. 

214.  Name. 
Terms,  i  „,  _ 

215.  Measure. 

I  216.  Solid  Contents. 
217.  TFoo(?. 

218.  Dry. 

219.  Liquid. 


Capacity. 


172 


DENOMINATE     NUMBERS. 


OUTLINE-Continued. 


MEASUHES, 


2,2,0.  Angular.  ■ 


335.  Time. 


f  221.  Vertex. 

222.  Circle. 

223.  Circumference 
,  224.  Measurement. 

226.  Solar  Day. 

227.  Sdar  Tear. 

228.  CivUDay. 

229.  Ztfop  Pear*. 


MISCELLAKEO  US. 


230.  Counting. 

231.  Paper. 

232.  Books. 

233.  Copying. 

234.  Shoemakers^. 


'  236.  Simple  or  Compound  JDenom- 
inate  to  Simple  Denominate. 

237.  Simple  Denominate  to  Simple 
or  Compound  Denominate. 


235.  REDUCTION.  - 


238. 


Denominate    Fractions    to 
Higher  or  Lower  Ternts, 


239.  Simple  Denominate  Fractions 

to  Simple  or  Compound  De- 
nominate Numbers. 

240.  A   Simple  or   Compound   De- 

nominate Numlter  to  a  Sim- 
ple Denominate  Fraction. 

241.  Finding  what  Part  One  De- 

nominate    Number     is     of 
Another. 


242.  ADDITION. 

243.  SUBTRACTION. 

244.  MULTIPLICATION, 
^  246.  DIVISION. 


CHAPTER    IV 


iix-^ 


IlEliaiOjMilMAT'El  MITiMBE'KS 


SECTION    I. 
DEFINITIONS. 

188.  Simple  Numbers  are  those  that  express  only 
one  kind, or  denomination.  Thus,  five,  fifteen  (meaning  five 
ones,  fifteen  ones),  five  apples,  fifteen  trees,  six  bushels,  three 
yards,  are  simple  numbers. 

189.  A  Shnjyle  Denominate  Number  is  a  sim- 
ple number  whose'  unit  is  used  as  a  measure  (191)  of  that 
number.  Thus,  5  dollars,  11  miles,  10  acres  are  denominate 
numbers,  because  1  dollar,  1  mile,  1  acre,  are  respectively 
exact  divisors,  or  measures  of  those  numbers. 

190.  A    Compound   Denominate   Nwinber, 

sometimes  called  a  Compound  Number ^  consists  of 
two  or  more  properly  related  simple  denominate  numbers 
taken  together.  Thus,  5  yards,  3  feet;  3  quarts,  2  pints,  1 
gill ;  1  mile,  7  furlongs ;  are  compound  denominate  numbers. 

191.  A  Measure  of  a  denominate  number  is  a  stand- 
ard tmit,  with  which  we  compare  the  number,  and  thus 
determine  its  numerical  value,  weight,  or  size.  Thus,  the 
measure  of  United  States  money  is  the  dollar  ( $ ) ;  of 
English  Money,  the  pound  ( £ ) ;  of  wood,  the  cord;  &c. ;  &c. 

113 


174 


TABLES. 


UNITED   STATES    COINS. 

GOLD. 


SILVER. 


BRONZE. 


NICKEL. 


SECTION    II. 


(I  TABLE  S^^ 


MONEY,  WEIGHTS  AND  MEASURES^^ 


7- 


-A 


MONEY. 


192.  Money  is  any  thing  used  as  medium  of  Com- 
merce; but  among  civilized  nations  is  usually  stamped 
metal,  called  coinSj  or  printed  bills  or  notes,  called  Paper 
Money, 

UNITED    STATES    3IONEY. 

193.  Federal^  or  United  States  Money ^  is  the 

legal  currency  of  the  United  States.     The  unit  of  this  cur- 
rency is  the  dollar. 

The  Currency  of  a  country  is  the  money  employed  in  the  com- 
merce of  that  country.     It  consists  of  both  iiaper  money  and  coin. 

A  M.int  is  a  place  where  money  is  coined. 

Bnllion  is  gold  and  silver  uncoined,  or  coined  gold  and  silver 
when  estimated  in  bulk  by  weight. 

TABLE. 


10  mills  (m.)  =  1  cent,  (f.) 
10  cents  =  1  dime,  (d.) 
10  dimes  =  1  dollar.  ($.) 
10  dollars      =  1  eagle.  (E.) 


E.   $    d.     ^     m. 

1  :=  10  =  100  =  1000  =  10000 

1=  10=  100=  1000 

1=   10=   100 

1=   10 


Each  of  these  denominations,  with  the  exception  of  mills,  is  repre- 
sented by  a  coin,  or  piece  of  stamped  metal,  issued  by  the  U.  S.  Mint, 
which,  in  addition  to  the  above,  issues  several  other  coins,  all  of  which 
are  represented  in  the  following  tables ; 

116 


176  TABLES. 

GOLD    COINS. 


2i  " 

f^^-^ 

.  .9     ' 

.05 

- 

.05 

« 

T.V.XgiX 

64.5 

3    " 

" 

.  .9     ' 

.05 

" 

.05 

" 

*' 

77.4 

5    " 

"    iE.. 

.  .9     ' 

.05 

" 

.05 

u 

" 

129. 

10  " 

«       E.. 

.  .9     ' 

'     .05 

ft 

.05 

f( 

« 

258. 

20  " 

"D.  E.. 

.  .9     ' 

.05 

K 

05 

M 

M 

516. 

SILVER    COINS. 

(Composed  of  .9  silver,  .1  copper.) 
Trade  Dollar  piece weight  27.22  -f  grammes  =  420       grains. 


Dollar  *♦ 

Haif-Dollar  " 
Quarter-Dollar  " 
Ten-Cent  " 

The  Trade  Dollar 
1 '.astern  countries. 


"       26.73+         '*         =412.5  '* 

"       12^  "         =  192.9  4-     " 

"         Q\  "         =    96.45+     " 

"         2|  "         =    38.58+     " 

is  intended  to  be  used  only  in  commerce  with 


COPPER   AND   NICKEL    COINS. 

(Composed  of  .75  copper,  .25  nickel.) 

Five-cent  piece. . .  .weight,  5  grammes  =  77.16  +  grains. 

Three-cent  "     "  30. 

One-cent      '*     95  copper,  .05  tin  and  zinc,  in  such  proportion 

as  may  be  ordered  by  the  director  of  the  mint ;  weight,  48  grains. 

The  Treasury  Department  of  the  United  States  issues  what  are 
called  «  United  States  Notes,"  of  $1,  $2,  $5,  |10,  $20,  $50,  $100,  $200, 
$500,  and  $1000  notes,  the  payment  of  all  of  which  is  secured  by  the 
United  States  Government. 

Mill  is  derived  from  mille  (1000) ;  Cent,  from  centvm  (100) ; 
Dime  is  from  the  French;  Dollar,  from  the  Danish  Daler,  or 
the  German  Thaler. 

The  symbol  $  is  probably  "  U  "  written  over  "  S,"  meaning  United 
States. 

CANABA   MO:SBY. 

194.  Canada  Money  is  the  legal  currency  of  the 
Dominion  of  Canada.    The  Unit  of  Measure  is  the  dollar. 


TABLES. 


177 


equal  in  value  to  the  United  States  gold  dollar.  The  other 
denominations  are  cents  and  mills,  and  have  the  same 
nominal  value  as  the  corresponding  denominations  of  U.  S. 
Money.     English  currency  is  also  much  used  in  Canada. 


Oold  Coins. 
Sovereign. 
Half  Sovereign. 


GOLD — 
SOVEREIGN. 


Siher  Coins.  Bronze  Coin. 

Fifty-cent  piece.  One-cent  piece. 

Twenty-five-cent  piece. 
Twenty-cent  piece. 
Ten-cent  piece. 
Five-cent  piece. 

ENGLISH  MONEY. 
COINS, 


SILVER. 


In  Great  Britain  the  gold  coins  are  22  carats  fine ;  that  is,  they  are 
■^  pure  gold  and  ^  alloy.  The  fineness  of  silver  is  estimated  by  the 
number  of  ounces  of  fine  silver  in  a  pound  Troy  of  metal.  Pure  silver 
is  12  fine  ;  and  silver  coins  are  lly^y  fine ;  that  is,  they  are  j^J  pure 
silver  and  y|^  alloy. 

195.  English  Money ,  sometimes  called  Sterlingf 
is  the  legal  currency  of  Great  Britain.  The  Unit  of  Meas- 
ure is  the  pound  (£)  =  $4.8665. 

TABLE, 

4  farthings  (far.  or  qr.)=l  penny d.  £    s.      d.      far. 

12  pence                           =1  shilling     .    .     .     .  s.  1=20=240=900 

20  shillings                     =1  pound,  or  sovereign  £  1=  12=  46 

21  shillings                     =1  guinea G,  1=    4 

13 


178 


DENOMI]S^ATE     KUMBEKS. 


Gold  Coins. 
Sovereign,  20s.  =  $4.8665. 
Half  "        10s.  =  $2.43335 
\^  gold,  and  -^^  alloy. 


Silver  Coins. 
Crown,    5s.  =  $1,216  +  . 
Half-  "      2Js.  =  $  .608  +  . 
Florin,         2s.  =  $  .487+. 
Shilling,    12d.  =  $.243+. 
Six-penny  piece  =  $  .12166  +  , 
Four-    "        "     =$  .0811+. 
Three-"        "     =  $  .06083  +  . 
f  ^  silver,  and  ^%  alloy. 
7s.  6d.  are  often  written  7/6 ;  5s.  8d.,  5/8  ;  &c.,  &c. 


Copper  Coins. 
?enny, =$.02027 ;  half -penny,  and 
farthing. 


GOLD. 


SILVER. 


GERMAN  MONEY. 
COINS. 

Many  of  the  old  coins  of 
Germany  are  still  in  use, 
notably  the  Silver  Thaler 
(Prussia),  valued  at  $.746, 
and  the  Silver  Oroschen, 
worth  2^^.  The  mark  was 
established  in  1872. 

Note. — In  estimating  the  value  of  coins,  it  is  only  the  fine  metal 
in  them  that  is  taken  into  consideration.  The  alloy  is  regarded  as 
valueless. 

196.  German  Money  is  the  currency  of  the  New 
German  Empire.  The  U?iit  of  Measure  is  the  "  mark " 
=  $.238.     Weight,  =  .3982  grammes. 


GERMAN    COINS. 

(.9  gold  and  .1  alloy.) 
Gold.— 20,  10,  and  5-mark  pieces  =  respectively  $4.76,  $2.38,  and 
$1.19. 

Silver.— 5,  2,  and  1-mark,  and  20-pfennig  pieces  =$1.19,  $.476,  $.238. 
and  $.0476. 

Copper, — Pfennig,  and  all  smaller  pieces. 

100  pfennigs  =  1  mark. 


TABLES. 


179 


GOLD. 


FRENCH    310NET. 
COINS. 

BKONZE. 


SrLVER. 


197.  French  Money  is  the  legal  currency  of  France. 
The  Unit  of  Measure  is  1  franc  =  ^.193.  It  is  founded  on 
the  decimal  system. 

TABLE. 

:  fr.      dc.         ct.  ms. 

10  millimes  (ms.)  =  1  centime. .  ct.  i  1  =  10  =  100  =  1000 
10  centimes  =  1  decime.  .  dc.  j  1  =    10  =    100 

10  decimes  =  1  franc.   .  .  fr.  i  1  =      10 

Millimes,  like  our  mills,  are  not  coined. 

Decimes,  like  our  dimes,  are  not  expressed;  thus,  6  fr.,  3  dc,  and 
7  ct.  are  written  Q.^^-21,  or  6.27  Ir.,  and  read  6  francs  and  27  centimes, 
or  6  and  27  hundredths  francs. 


Gold. 

100-franc  piece. 

40    " 

20    " 

10    " 

5    " 


FRENCH   COINS. 

Silver. 

5-franc  piece. 

2    " 

1    " 
50-centimes. 
25       " 


Bronze. 
10  centime  pie  )e. 
5      '• 
2      " 
1      " 


The  Napoleon  is  a  20  franc  piece  =  $3.86. 
The  value  of  the  franc  in  estimating  duties,  is  $.193 
value  is  about  the  same. 

The  coins  of  France  are  .9  pure  metal  and  .1  alloy. 


its  intrinsic 


180  DENOMINATE     NUMBERS. 


WEIGHT. 

198.  Weight  is  the  measure  of  the  force  of  gravity. 

Gravity  is  the  teDdency  of  bodies  to  fall  toward  some  centre. 
Bodies  on  the  earth  tend  toward  the  earth's  centre. 

Weights  are  divided,  with  respect  to  the  use  to  which  they  are 
applied,  into  Troy  Weight,  Apothecaries'  Weight,  and  Avoirdupois 
Weight. 

TROY  WEIGHT. 

199.  Troy  Weight  is  used  in  weighing  precious  metals 
and  jewels,  and  in  philosophical  experiments. 

The  Unit  of  Measure  is  one  pound  =  5760  grains. 


TABLE, 


24:  grains  (gr.)  =  1  pennyweight  pwt. 
20  pennyweights  =  1  ounce .  .  .  oz. 
12  ounces  =  1  pound      .    .    lb. 


lb.    oz.     pwt.       gr. 
1  =  12  r=  240  =  5700 
1=  20=  480 
1=     24 


The  Troy  povnd  is  the  standard  of  weight  at  the  U.  S.  Mint. 

A  Carat  is  a  weight  used  to  weigh  diamonds  and  precious  stones. 
It  is  equal  to  3.2  grains  Troy,  or  4  carat  grains. 

Carat  is  also  a  word  used  to  indicate  the  fineness  of  gold.  Thus, 
gold  when  24  carats  fine,  is  all  gold ;  when  15  carats  fine,  it  is  ^  or 
.625  gold,  and  ^  or  .375  alloy. 

Troy  Weight  received  its  name  from  Troyes,  a  town  in  France  where 
this  weight  was  first  used  in  Europe.  It  is  said  to  have  been  brought 
from  Cairo,  in  Egypt,  during  the  crusades,  and  adopted  by  the  gold- 
smiths, many  of  whom  resided  at  Troyes. 

The  old  silver  penny  of  England  was  used  as  a  weight,  hence  the 
term  'pennyweight. 

The  term  grain  originated  in  the  custom  of  using  a  number  of 
grains  of  wheat  as  the  weight  of  a  penny.  At  first  32,  and  afterwards 
24  made  a  pennyweight.  The  grains  were  taken  from  the  middle  of 
the  ear  and  well  dried. 


ARaTjHEGARIEj  WEIGHT  AN  a  M  ENURES. 


^yi 


1-^ 


T^ 


200.  Apothecaries^  Weight  is  another  form  of  Troy 
"V^'eight,  and  is  used  in  compounding  medicines  when  in  a 
d-y,  or  solid  form.  The  Unit  of  Measure  is  one  pound  = 
5760  grains. 

TABLE, 

i  fc       !        3        3  gr. 

20  grains  (gr.)  =  l  scruple.  .    3    ;  1=12=96=288  =  5760 

3  scruples       =ldram.  .  .    3    i  1=  8=  24=  480 

8  drams  =1  ounce  .  .    1     i  1=     3=     60 

12  ounces         =1  pound  .  .    lb   J  1=     ^0 

181 


182  DEN^OMINATE     NUMBEES. 

The  prescriptions  of  physicians  are  written  in  Roman  notation,  a 
small  j  being  used  for  small  i  when  final.  Thus,  2  scruples  are  writ- 
ten 3ij  ;  7  drams,   3  vij  ;  12  ounces,  §  xij. 

Apothecaries'  Fluid  Measures  are  those  used  in  compound- 
ing medicines  when  in  a  liquid  foi-m. 

TABLE, 

60  minims  (or  drops)  =  1  fluid  drachm f  3  . 

8  fluid  drachms  =  1  fluid  ounce f  5  . 

16  fluid  ounces  =  1  pint 0. 

8  pints  ==  1  gallon Cong. 

Cong.  1  =  0.  8  ==  f  5  128  =  f  3  1024  =z  m  61440. 

(1.)  Cong.,  from  the  Latin  congius,  means  gallon. 
(2.)  0.,  from  the  Latin  octarius,  means  one-eigliili. 
(3.)  The  minim  is  atout  equal  to  a  drop  of  water. 

The  following  measures,  from  vessels  in  common  use,  are  sometimes 
used: 


4  tea-spoons     =  1  table-spoon. 

2  table-spoons  =  1  ounce. 

2  ounces  =  1  wine-glass. 


2  wine-glasses  =  1  tea-cup. 
4  tea-cups        =  1  pint. 


In  AjKJthecaries'  Weight,  the  pound,  ounce  and  grain  are  the 
same  as  in  Troy  Weight,  but  the  ounce  is  differently  divided 
It  is  a  modification  of  Troy  Weight  made  by  the  apothecaries  of  early 
tames  and  applied  to  their  own  special  pursuit. 

The  peculiar  characters  used  in  this  weight  are  supposed  to  have 
been  taken  from  inscriptions  upon  the  ancient  monuments  of  Egypt. 

It  was  the  design  of  apothecaries  and  physicians  carefully  to  con- 
ceal from  others  all  knowledge  of  the  mixtures  given  as  medicines,  and 
hence  the  articles  composing  medicines  were  named  in  Latin,  and 
arbitrary  signs  were  used  to  express  the  quantity. 

The  Standard  Unit  of  Weight  adopted  by  the  United  States  is  the 
Troy  Pound  of  the  United  States  Mint.  It  is  equal  to  22.794422  cubic 
inches  of  distilled  water  at  its  maximum  density,  the  barometei 
standing  at  30  inches. 


201.  Avoirdupois  Weight  is  the  common  weight  of 
commerce.  It  is  used  in  weighing  all  articles  sold  by  weight 
except  precious  metals,  precious  stones,  liquids,  &c.  The 
Unit  of  Measure  is  one  pound =7000  grains. 


TABLE, 

16  drams  (dr.)  =  1  ounce      .    .    oz. 

16  ounces         =1  pound     .    .     lb. 

100  pounds  =1  hundredweight  cwt. 

20  hundredweight =1  ton  .    .     T. 


T.  cwt.      lb.         oz. 
1=20=2000=32000 
1=  100=  1600 
1=       16 

183 


184  DENOMINATE     NUMBERS. 

Formerly  in  this  weight  there  were  2340  pounds  in  a  ton,  and 
1 12  pounds  in  a  hundredweight,  which  was  divided  into  four  quarters 
of  28  pounds  each.  These  are  now  called  the  long  ion  and  the  long 
hundred,  and  they  are  still  the  authorized  weight  of  England,  but  in 
this  country  they  are  not  used,  except  at  the  custom-houses  for 
weighing  goods  imported  from  Great  Britain,  and  sometimes  for 
weighing  coal,  iron,  and  plaster. 

The  following  is  the  English  Avoirdupois  Table. 


16  drams       —  1  ounce 

T.  cwt.  qr.      lb.        oz. 

16  ounces     =  1  pound 

1=20=80=2240=35840 

28  pounds     =  1  quarter. 

1=  4=  112=  1792 

4  quarters  =  1  hundred. 

1=    28=    448 

20  hundred  =  1  ton. 

1=      16 

Under  the  head  of  Avoirdupois,  the  following  weights  are  in  com- 
mon use : 

100  lb.  Flour  or  Grain  =  1  Cental. 

100  lb.  Nails  =  1  Keg. 

100  lb.  Raisins  =  1  Cask. 

100  lb.  Dry  Fish  =  1  Quintal. 

196  lb.  Flour  =  1  Barrel. 

200  lb.  Pork  or  Beef  =  1  Barrel. 
280  lb.  Salt  at  U.  S.  Salt  works  =  1  Barrel. 

1000  oz.  Water  =  1  Cubic  foot. 

The  origin  of  the  term  Avoir  du  pois  is  not  certainly  known  and 
different  theories  are  given  to  account  for  its  adoption  and  use  : 

1.  That  it  is  from  the  Norman  term  Avoir  du  poids,  which 
signifies  goods  or  chattels  of  weight. 

2.  That  it  is  from  the  old  French  Aver  de  pes,  goods  of  weight. 

3.  That  it  is  from  an  old  French  verb  Averer,  to  verify. 

4.  That  it  is  from  the  French  Avoir  du  poids,  to  have  weight. 

The  following  table  shows  the  weight  of  a  bushel  of  the  commodi- 
ties named  in  the  left  hand  column,  as  fixed  by  statute  in  the  States 
named  at  the  top : 


TABLES. 


185 


WEIGHT    OP    A    BUSHEL. 


i 

. 

i 

3 

a 

i 

&2 

o3 

1 

-a '9 

a '.a 

1 

O 

i 
1 

".S 

Barley       

50 
40 

i: 

52 
32 

45 

56 

56 

48  48  48  48 
60  60  60  60 
14141414 

52  50^52" 

-'-1  — 
46  46  46 

60^60  60  go" 

a4  25  24 

33,33  33, 

55  56  56  56 

44  4444  44 

32 

46 
46 

56 

50 

30 
52 

56 
50 

48  48  48 

_ 

30 
60 

48 

50 
64 

55 
56 

_ 
30 

60 
56 

48  48 

46 

m 

28 
28 

56 

34 

60 
56 

47 
60 
14 
48 
46 
62 
24 
83 
56 
44 
56 
70 

^ 
76 

^ 

57 

56 

56 

^5 

50 

50 
60 

50 

46 
46 

56 

32 

60 
56 

60 

45 

42 

60 
28 
28 

56 

36 
50 

m 

56 

48  4S 

:i: 

42  42 

60 
14 
62 

24 
i 
56 
44 
52 

80 
32 
57 
60 
56 

50 
45 

60 
60 

! 

58 

32 

60 
56 

56 

44 

— 

60 

56 

32 

56 

_ 

Beans -. 

56 

32 
32 

_ 

50 
30 
60 
50 

_ 
42 

60 

28 
28 
56 

56 

32 

60 
56 

46 

Blue-Grass  Seed  . 

Buckwheat 

50 

Castor  Beans 

Clover  Sfied 

60 

60 

Dried  Apples  .... 

28 
56 

32 
56 

28 

S 

56 

Dried  Peaches.... 

Flax  Seed 

Hemp  Seed 

Indian  Com 

56 
70 

80 
32 
57 

56  56  56 

54 

Indian  Com  in  ear 

68 
50 

70 

50 

mi 

57 
56 

50 

Indian  Com  Meal. 

Bituminous  Coal.. 

70 

* 

Oats 

32  32 

1 

Onions... 

48 

57 

Potatoes 

54 

60 
56 

- 

60 
56 

50 
45 

60 

Rye 

56  56 

III 
50  50 

1^^ 

Rye  Meal       .  .  . 

Salt 

1 

1 

Timothy  Seed.... 

1 

Wheat 

|60 

^ 

60 

60 

60  BO 

60 

60 

60 

60 

60 

60 

60 

60rin 

60^0 

60 

GOCJi 

jW 

In  Pennsylvania,  85  lb.  coarse,  70  lb.  ground,  or  62  lb.  fine  salt  make 
1  bushel,  and  280  lb.  =  1  bbl. 

Comparison  of  Avoirdupois,  Apothecaries^  and  Troy  Weiglit. 

7000  grains  =  1  lb.  Avoir.       43 7|^  grains     —  1  oz.  Avoir. 

5760      "      =1  lb.  Troy.        480      «  =1  oz.  Troy. 

5760      "      =1  lb.  Apoth.      144  lb.  Avoir.  =  175  lb.  Troy. 


M 


LON^,  SQUARE:,  Qliil(;,.W.QQQj  ANQj 
s.         SURVEYORS?  MEASURESj    '      d 


■^^ 


^1^ 


202.  Long,  or  Linear  Measure^  is  used  in 
measuring  distances,  or  anything  that  has  length.  The 
U?2it  of  Measure  may  be  1  mile,  1  rod,  1  yard,  &c.  A 
Line  has  but  one  dimension— /ew^^A. 

Linear  measure  is  so  called  because  it  may  be  taken  with  a  line  ;  it 
includes  Long  Measure,  Cloth  Measure,  and  Surveyors'  Linear  Measure, 

186 


mi.  fur.      rd.  ft.  in. 

1  =:8  =  320  =  5280  =  63360 

1  =    40  =    660  =    7920 

1  =  16.5  =z     198 

1=       12 


TABLES.  187 

TABLE. 

12  inches  (in.)  =1  foot  .  .   ft. 

3  feet  =  1  yard  . .  yd. 

5J  yards,  or  )^^^^^  .  rd. 

16i  feet  ) 

40  rods  =lfur'g  .  fur. 

8furlongs,or)       ^^.j^^^^. 

320  rods         ) 

60  ffeosraphic  miles,  or )       ^  ^  . ,     -r^      ,  ,  ^ 

nr^^n    L  .\       '1  C  ='^  degree  on  the  Earth's  Equator. 

69.16  statute  miles  )  ^  ^ 

3.60  degrees  =  Circumference  of  the  Earth. 

Since  the  earth  is  not  a  perfect  sphere,  the  degrees  of  latitude  are 
not  of  equal  length.  Their  average  length,  69.16  miles,  is  therefore 
adopted.  The  geographic  mile  =  1.15  statute  miles,  very  nearly.  The 
geographic  mile  is  adopted  by  navigators  on  account  of  its  coJivenience, 
1  mile  corresponding  to  1  minute  of  arc. 

doth  Measure  has  but  one  denomination,  the  ya/rd,  which  is 
divided  into  Iwlves,  quarters,  eighths,  sixteenths,  &c. 

The  following  also  come  under  the  head  of  Long  Measure : 

6  feet  =  1  fathom  (used  in  measuring  depths  at  sea). 

128  fathoms      =  1  cable's  length. 

3  geo.  miles  =  1  league  =  3.45  statute  miles. 

4  inches        =  1  hand  (used  in  measuring  height  of  horses). 
21.888  inches        =  1  cubit  (mentioned  in  the  Bible). 

3  inches        =  1  palm. 

12  lines  =  1  inch. 

3  feet  —  1  pace. 

A  Knot,  used  in  measuring  the  speed  of  vessels  at  sea,  is  equal  to  a 
nautical  mile. 

203.  Surveyors'^  Linear  Measure  is  used  in 
measuring  roads,  and  boundaries  of  land.  The  Unit  of 
Measure  is  1  chain  =  66  feet. 


188 


DENOMINATE     KUMBEUS, 


TABLE, 


7.92  inches  =  1  link  .  . .    1. 

100  links    =  1  chain  .  .  ch. 

80  chains  =  1  mile  . . .  mi. 


mi.       ch. 
1  =  80 

1 


1.  in. 

8000  =  63360. 
100  =      792. 
1  =      7.92. 


Tlie  pole  {  =  25  links  =  16|  feet)  is  rarely  used,  chains  bein»  writ- 
ten as  a  whole  number  and  links  as  a  decimal. 

Engineers  usually  use  a  chain  divided  into  feet,  or  a  tape  measure 
divided  into  feet  and  inches,  or  into  feet  and  tenths  and  hundredths 
of  a  foot. 

204.  Square  Measure  is  used  in  finding  areas  of 
surfaces,  or  of  that  which  has  length  and  breadth  without 
thickness.  Its  Units  of  Measure,  usually,  are  square  surfaces, 
each  of  whose  sides  is  a  linear  unit  of  the  same  name. 

205.  An  Angle  is  the  opening,  or  divergence,  of  two 
lines  which  meet.    Thus, 


A  B  and  A  C  meet  at  the  point  A, 
forming  an  angle  A,  or  B  A  C 

The  lines  A  B  and  A  C  are  called 
the  sides  of  the  angle. 


206.  When  two  lines  meet  so  that  the  adjacent  angles 
formed  by  them  are  equal,  the  angles  are  called  Right 
Angles,  And  the  lines  are  said  to  be  perpendicular  to 
each  other.    Thus, 


If  A  B  and  C  D  meet  so  as 
to  make  the  angle  A  D  C  = 
the  angle  B-D  C,  then  these 
angles  are  right  angles,  and 
the  lines  A  B  and  C  D  are  per 
p&ndicvXar  to  eacb  other. 


T A  BLES 


189 


201,  A  figure  bounded  hj  four  straight  anes,  and  having 
four  right  angles  is  called  a  Hectangle,  and  is  said  to 
have  two  dimensions,  length  and  breadth. 

208.  If  the  sides  are  at  the  same  time  equal  in  length 
then  the  figure  is  called  a  Square,    Thus, 

The  figure  ABCD  represents  a  Square. 
If  each  of  the  sides  AB,  BC,  CD,  and  BA 
is  a  linear  inch,  then  the  square  is  a 
square  inch  ;  if  each  side  is  a  linear  foot, 
the  square  is  a  square  foot  ;  if  each  side  is 
a  linear  yard,  the  square  is  a  square  yard  ; 
and  in  like  manner  we  have  a  square  rod, 
a  square  mile,  &c. ;  and  also  a  square  link 
and  a  square  chain. 

209.  Each  square  may  be  used  to  measure  a  surface, 
and  the  number  of  these  squares  that  equal  any  surface  is 
called  the  Measure  of  that  surface.    Thus, 


FiGUBE  1. 


Figure  2.    Figure  3 


S  X^gyy-  Z.OA/G. 

IlilWpiilffipiiilliplBll 


"""lllillllilll 


jlli;iiii,::tiili 


!!||ll, 


I'll 


'""iilil; 


\ 


Siii 


"lllll,,  ••"illllll 


//"oor. 


nil 


% 


/Foot. 


(4  sq.  ft.  X  5  =  20  sq.  ft.) 

In  Fig.  2  there  are  4  squares  each  —  in  size  to  Fig.  3  which  we 
will  call  a  square  foot ;  hence  there  are  4  square  feet  in  Fig.  S. 

Since  Figure  1  contains  5  rows,  each  of  which  is  =  in  size  to 
Fig.  2,  there  are  20  square  feet  in  Fig.  1. 


190  DENOMIN^ATE    NUMBEnS. 

It  will  be  observed  that  this  is  the  same  result  as  would  be  found 
by  multiplying  the  length  and  breadth  of  Fig.  1  together.  We, 
therefore,  conclude  that 

310.  The  Area  of  any  riglit-angled  four-sided  figure 
will  contain  as  many  square  units,  as  there  are  units  in 
the  product  of  the  number  of  linear  units  of  the  same  name 
in  its  length  and  breadth. 

We,  therefore,  readily  see  how  the  table  of  Square  Measure  is 
derived  in  great  part  from  that  of  Long  Measure. 

TABLE. 

144  (=12  X  12)  square  inches  (sq.  in.)  =    1  square  foot sq.  ft. 

9  (=  3  X   3)  square  feet  =    1  square  yard sq.  yd, 

30}  (=3  5|  X   5i)  square  yards,  or  )  ,  (  '^^  "  *  '  *  ^-  ^^' 

272i  (=16i  X  16i)  square  feet,         (    "        '^''^''  )  ^T  '  ""''^'^ 

;  (  pole  .  .  .  sq.  p. 

40  poles  =    1  rood R. 

4  roods  =    1  acre A. 

640  acres  =    1  square  mile sq.  mi. 

311.  Surveyors^  Square  Measure  is  used  in 
obtaining  the  area  or  contents  of  land.     The  acre  is  the 

Unit  of  Measure. 

TABLE. 

10000  (=100  X 100)  square  links  (sq.  1.)     =     1  square  chain  .  .  sq.  ch. 

10  square  chains  =     1  acre A. 

640  acres  =     1  square  mile  .  .  .  sq.  mi. 

625  square  links  =  1  square  pole,  and  16  sq.  p.  =  1  sq.  ch. 

One  sq.  mi.  is  called  a  section,  and  36  sections  placed  so  as  to  form 
a  square  are  called  a  toumship. 

This  applies  only  to  the  new  States  which  have  been  laid  out  by 
authority  of  the  General  Government.  The  older  States  are  not  reg- 
ularly laid  out,  and  hence  their  townships  are  not  uniform  in  size  and 
shape. 

A  township  is  a  division  of  a  county  made  for  convenience  in  hold- 
ing elections.  There  must  be  at  least  one  voting  place  in  each  town- 
ship. 


TABLES. 


191 


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In  1802, the  following  very  convenient  method  of  laying  out  Gov» 
ernment  lands  was  adopted  by  Col.  Jared  Mansfield,  at  that  time 
Surveyor-General  of  the  North- Western  Territories : 

The  country  to  be  surveyed  is  divided  by  North-and-South  lines 
(meridians),  six  miles  apart ;  these  are  intersected  at  right  angles  by 
East-and-West  lines,  six  miles  apart,  thus  dividing  the  land  into 
squares,  called  Townships,  6  miles  each  way, 

A  Rarif/a  is  a  line  of  Townships  running  North  and  South,  and  is 
known  by  its  number  East  or  West  of  some  given  North-and-South 
line,  called  a  principal  meridian. 

The  Townships  are  also  numbered  North  and  South  from  some 
established  East-and-West  lines,  called  parallels. 

Townships  are  also  divided  into  square  miles,  called  Sections^  and 
are  numbered,  beginning  at  the  N.  E.  corner,  as  shown  in  the  diagram. 


192  DENOMINATE     NUMBERS. 

Each  Section  is  again  sub-divided  by  meridians  and  parallels  into 
Half- Sections,  Quarter- Sections,  Half-  Quarter-Sec- 
tions, ^xA  Quarter-Quarter-Sections.    Thus, 

Section  9  is  divided  into  four  quarter-sections,  and  the  S.  E. 
quarter  is  again  sub-divided  into  four  equal  parts  designated  by  the 
letters  p,  q,  r,  s,  and  named  as  follows  :  p  is  the  W,  |  of  S.  E.  ^  of 
Section  9 ;  g  is  the  E.  |  of  W.  |  of  S.  E.  \  of  Section  9 ;  /•  is  the  W.  ^ 
of  the  E.  ^  of  S.  E.  ^  of  Section  9 ;  and  8  is  the  E.  |  of  same  \  section. 

Section  11  ia  also  divided,  some  of  the  divisions  being  designated 
by  the  letters  a,  b,  c,  d,m,  and  A,  and  named  as  follows : 

a,  N.  E.  \  of  N.  E.  ^,  (or  N.  i  of  E.   ^  of  N.  E.  ^). 

b,  N.  W.  ^  of  N.  E.  \,  (or  N.  i  of  W.  ^  of  N.  E.  \). 

c,  S.  W.  \  of  N.  E.  ^,  (or  S.  ^  of  W.  ^  of  N.  E.  ]). 

d,  S.  E.    I  of  N.  E.  i,  (or  S.  i  of  E.    ^  of  N.  E.  \). 
A,  West  ^  of  Section  11,  containing  330  acres. 

m,  S.  E.    i  of  Sec.  11. 
Section  21  is  divided  into  parts  designated  B,  e,  f,  g,  and  h,  and 
named  as  follows  : 

B,   N.   ^  of  Section  21  ;  320  acres. 
e,     E.    i  of  S.  E.   ^  of  Sec.  21 ;  80  acres. 
/,    W.  ^  of  S.  E.    J  of  Sec.  21 ;  80     " 
g,     N.  \  of  S.  W.  \  of  Sec.  21 ;  80     " 
h,     S.   i  of  S.  W.  i  of  Sec.  21 ;  80     " 

213.  Solid,  or  Cubic  Measure,  is  used  in  measuring 
'contents  or  volume  of  solids.    Its  ineasuring  units  are  cubes, 

213.  A  Cube  is  a  body  bounded  by  six  equal  squares. 
Therefore  its  length,  width,  and  thickness  are  equal.  It  is 
said  to  have  three  dimensions,  namely,  length,  breadth,  and 
thichness. 

214.  A  cube  takes  its  Name  from  the  length  of  one  of 
its  edges.  Thus  a  cubic  inch  is  a  cube,  each  of  whose  edges 
is  an  inch  long  ;  a  cubic  foot  is  a  cube,  each  of  whose  edges 
is  a  foot  long ;  &c. 

215.  The  solidity,  or  contents,  of  a  body  is  the  quantity 
of  space  which  it  occupies ;  that  is,  its  bulk,  or  volume  j  and 


TABLES 


193 


the  number  of  cubes  that  are  equal  to  the  solidity  of  any 
body  is  the  Measure  of  that  body.    Thus, 


Fig.  1. 


Fig.  2. 


/fr.X/rrX/fr.=  I  Cub/c  Foot. 
£jrr.XSAT.X6rT.'^^^  Cubic  Fser. 


(2x2x6  =  24  cu.  ft.) 


(1ft.  X  1ft.  X  lft.  =  lcu.  ft.) 


Fig.  1  represents  in  the  bottom  tier  2  rows  of  blocks,  each  block 
being  =  in  size  to  Fig.  2,  and  6  blocks  in  eacli  row,  or  2  x  6  =  12 
blocks  in  the  lower  tier. 

In  the  upper  tier  there  are  also  2  rows  with  6  blocks  in  each  row, 
or  2  x  6  =  12  blocks  in  the  upper  tier.  In  both  tiers  there  are 
2  x  (2  x  6)  =  24  blocks.  Therefore  we  measure  Fig.  1  by  Fig.  2,  and 
find  the  former  to  be  24  times  the  latter. 

If  Fig.  2  represents  a  cubic  inch,  Fig.  1  is  24  cvMc  inches  ;  if  Fig.  2 
is  a  culm  foot.  Fig.  1  is  24  cubic  feet ;  if  Fig.  2  is  a  cubic  yard^  cubic 
rod,  cubic  mile,  or  cubic  chain,  then  Fig.  1  is  respectively  24  cubic 
ya/rds,  24  cubic  rods,  24  cubic  miles,  or  24  cubic  chains. 

Had  we  multiplied  together  the  represented  breadth,  height,  and 
length  of  Fig.  1,  we  should  have  obtained  2x2x6  =  24,  the  same 
result  as  before.     We,  therefore,  conclude  that 

216.  To  obtain  the  solid  contents  of  any  rectangular 
body,  we  multiply  together  its  length,  breadth,  and  thickness, 
or  its  three  dimensions. 

13 


194  DEKOMIKATE    NUMBERS. 

TABLE. 

1728  (=12  X 12  X 12)  cubic  inches  (cu.  in.)  =  1  cubic  foot    .    .    cu.  ft. 

27  (=  3  X   3  X   3)  cubic  feet  =  1  cubic  yard   .     .     cu.  yd. 

24f  (=16^  X  U  X  1)  cubic  feet  =  1  perch  of  stone. 

40  cu.  ft.  in  the  U.  S.  and  42  cu.  ft.  in  England  =  1  shipping  ton. 

100  cubic  ft.  =  1  register  ton,  used  in  estimating  the  tonnage  of 
vessela 

217.  Wood  Measure, 

Wood  measure  is  used  for  measuring  wood  prepared  to  be  used 
as  f  ueL 

The  unit  of  measure  is  the  cord. 

TABLE. 

16  cu.  ft.  =  1  cord  foot cd.  ft. 

8  cord  feet,  or 


-„o        x^i.  ^  =  1  cord cd 

128  en.  ft. 

A  cord  of  wood  is  a  pile  of  wood  8  feet  long,  4  feet  wide,  and  4  feet 
high.     8x4x4=  128  cu   ft. 

A  cord  foot  is  a  pile  of  wood  4  feet  high,  4  feet  wide,  and  1  foot 
thick.    4x4x1  =  16  cu.  ft. 

Note  on  Meabhres.— Square  and  Cubic  Measures  (204-217) 
are  all  derived  from  Long  Measure  (202). 

The  Long  Measure  standard  is  obtained  in  this  way :  Theory,  con- 
firmed by  careful  experiments,  proves  that  a  pendulum  of  a  given 
length  will  make  a  single  vibration,  that  is,  will  swing  back  and 
forth  once,  in  a  portion  of  time,  which  is  invariable  at  the  same 
place.  It  has  been  ascertained  also  that  at  the  sea  level  in  London,  a 
pendulum  that  will,  in  a  vacuum,  vibrate  once  in  a  second,  is  nearly 
39.1393  inches  in  length.  This  pendulum,  then,  is  the  basis  of  the 
Long  Measure  standard,  ^^.f  f^^  of  it  is  called  a  yard ;  and  the  yard 
is  subdivided,  or  multiplied,  to  produce  the  other  measures  in  use. 

To  determine  measures  of  weight  from  Long  Measure,,  another 
element,  water,  is  used ;  and  22.79  +  cubic  inches  of  pure  water,  at  its 
greatest  density,  with  the  barometer  at  30  inches,  is  agreed  upon  as 
a  pound  Troy ;  and  having  the  pound  Troy,  it  is  an  easy  matter  to 
find  any  number  of  greater  or  less  units  of  weight. 

In  the  French,  or  metric  system,  measures  are  deduced  from  the 
metre,  whose  length  is  39.37079  inches,  which  is  nearly  one  ten- 
millioiith  of  the  distance  from  the  equator  to  either  pole. 


bu.      pk.       qt.        pt. 

qt. 

1  =  4  =  32  =  64 

pk. 

1=    8  =  16 

bu. 

1=2 

218.  Dry  Measure  is  used  in  measuring  the  bulk 

of  solid  matter  in  a  somewhat  divided  state;  such  as  fruit, 

grain,  salt,  ashes,  coal,  lime,  &c.     The  unit  of  measure  is  the 

bushel, 

TABLE. 

2  pints  (pt.)  =  1  quart  .  . 
8  quarts  =  1  peck  .  . 
4  pecks  =  1  bushel. 

The  standard  bushel  of  the  United  States  contains  2150.42  cubic 
inches.  It  is  conveniently  represented  by  a  cylindrical  vessel  18|- 
inches  in  diameter,  and  8  inches  deep.  It  is  the  old  English  or  Win- 
chester bushel,  and  equals  77.6274  lb.  distilled  water  at  its  greatest 
density.     1  pint  dry  measure  =  nearly  1 1  pt.  liquid  measure. 

The  standard  bushel  of  Great  Britain  contains  2218.192  cubic  inches. 

In  measuring  small  seeds,  beans,  peas,  oats,  rye,  wheat,  &c.,  the 
measure  must  be  enen  full.  In  measuring  bituminous  coal,  lime,  the 
larger  fruits,  &c.,  the  measure  must  be  heaped,  or  contain  f  of  2150.42 
I.  in. 

195 


219.  lAquid  Measure  is  used  in  measuring  wines, 
oils,  milk,  molasses,  brandy,  vinegar,  &c.,  &c.  The  Unit  of 
Measure  is  the  gallon. 

TABLE, 

gal.      qt.      pt.        gi. 

.  pt. 
.  qt. 
•  gal. 


4  gills  (gi.)  =  1  pint   , 
2  pints         =  1  quart 
4  quarts       =  1  gallon 


1  ==  4  r=  8  =  33 

1  =  2=8 

1=4 


The  gallon  contains  231  cubic  inches,  and  is  the  same  as  the  old 
English  Wine  Gallon,  and  is  equal  to  8.338  lb.  distilled  water  at  its 
greatest  density. 

In  calculating  the  capacity  of  cisterns,  reservoirs,  &c.,  a  barrel  is 
31.5  gal.,  and  63  gal.  are  called  a  hogshead,  though  barrels  and  hogs- 
heads as  used  in  commerce  are  of  different  capacities,  which  are  de- 
termined by  gauging. 
196 


TABLES.  197 

ANGVLAB  OR  CIRCULAR  MEASURE. 

220.  Circular  Pleasure  is  used  in  measuring  an- 
gles, arcs  of  circles,  latitude  and  longitude,  and  the  motions 
of  the  heavenly  bodies,  &c. 

The  Unit  of  Measure  is  the  degree,  which  is  ^^  part  of 
the  circumference  of  a  circle. 

An  Angle  is  the  difference  in  direction  of  two  lines  which 
meet  (205). 

221.  The  Vertex  of  an  angle  is  the  point  at  which  its 
sides  meet. 

The  Sides  of  an  angle  are  the  lines  which  form  it. 

The  kinds  of  angles  are  Right  Angles,  Acute  Angles,  and  Obtuse 
Angles. 

A  Right  Angle  is  the  angle  formed  by  two  straight  lines 
perpendicular  to  each  other  (206). 

An  Acute  Angle  is  one  which  is  less  than  a  right  angle. 

An  Obtuse  Angle  is  one  which  is  greater  than  a  right  angle. 

222.  A  Circle  is  a  plane  figure  bounded  by  a  curved 
line,  every  part  of  which  is  equally  distant  from  a  point 
within  called  the  centre. 

223.  The  Circumference  of  a  circle  is  the  line 
which  bounds  it.  Any  part  of  the  circumference  is  called 
an  Arc. 

The  Diameter  of  a  circle  is  a  line  which  passes  through 
the  centre  and  terminates  in  the  circumference. 

The  Radius  is  a  line  extending  from  the  centre  to  the 
circumference. 

224.  An  angle  is  measured  by  making  its  vertex  the 
centre  of  a  circle  and  reckoning  the  arc  included  between 
its  sides.  For  this  purpose  the  circumference  is  divided  as 
in  the  table. 


1^ 


DENOMINATE      NUMBERS. 


TABLE. 

60  seconds  of  arc  (")       =1  minute ', 

60  minutes  "    "  =1  degree    °. 

30  degrees    "     "  =1  sign S.  ' 

12  signs,  or  360  degrees  =  1  circumference  .  .  .  .  C. 
90  degrees  of  arc  =  1  quadrant  (quad.)  or  right 

angle  (r.  a.). 

C.       S.  °  '  " 

1  =  12  =  360  =  21600  =z  1296000. 

1  =  30  =  1800  =  108000. 

1  =   60  =   3600. 

1  =     60. 


ILLUSTBATION. 


The  lines  C  D  and 
C  A  include  a  quar- 
ter of  the  circumfer- 
ence between  them ; 
hence  their  differ- 
ence of  direction  is 
i  of  360° =90°.  Such 
an  angle  is  called  a 
right  angle  (206), 
and  the  measuring 
arc  is  called  a  quad- 
rant. The  lines  C  D 
and  C  A  are  said  to 
be  perpendicular  to 
each  other  (206). 

The  lines  C  B  and 
CE  include  one-sixth 
of   a  circumference 

between  them;  hence  their  difference  of  direction  is  ^  of  360° =60°, 
and  the  measuring  arc  is  called  a  sextant. 

An  arc  measuring:  |^  of  a  circumference  is  called  an  octant. 
All  angles  less  than  90°  are  called  acute  angles,  and  those  greater 
are  called  obtuse  angles. 


225.  Time  is  a  measurable  portion  of  duration. 

22G,  The  natural  measure  of  Time  is  the  Solar  I>ayf 
or  the  interval  of  time  between  two  successive  passages  of 
the  sun  07er  the  same  meridian  ;  but,  as  these  intervals  are 
of  unequal  length,  the  True  Unit  of  Measure  is  a  mean  of 
those  intervals,  and  is  called  a  Mean  Solar  Day ;  and  is  24 
hours  long. 

227.  A  Solar  Year  is  the  time  in  which  the  earth 
makes  one  revolution  round  the  sun. 

199 


200  DE  NOMINATE  NUMBERS. 

228.  A  Civil  Day  is  in  length  the  same  as  a  mean 
solar  day ;  that  is,  24  hours,  and  in  most  countries  begins 
and  ends  at  midnight. 

TABLE. 
60  seconds  (sec.)  =  1  minute  ....    min. 

60  minutes  =  1  hour hr. 

24  hours  =  1  day da. 

7  days  =  1  week wk. 

12  calendar  months  =  1  year yr. 

365  days  (or  52  wk.  1  da.)        =1  common  year. 

366  days  (or  52  wk.  2  da.)        =1  leap  year. 
365  da.  5  hr.  48  min.  49.7  sec.  =  1  solar  year. 
100  years  =  1  century. 

By  an  inspection  of  the  Table  the  pupil  will  see  that  a  solar  year  is 
5  hr.  48  min.  49.7  sec.  longer  than  a  common  year ;  and  that  4  solar 
years  are  =  4  common  years  +  23  hr.  15  min.  18.8.  sec.  To  make  the 
Solar  and  Common  year  correspond,  1  day  is  added  to  every  fourth 
common  year,  or  to  every  year  exactly  divisible  by  4 ;  and  that  year  ip 
called  a  leap  year.  But  by  adding  1  day  every  4  years,  too  much  is 
added  by  the  difference  between  24  hrs.  and  23  hr.  15  min,  18.8  sec, 
or  at  the  rate  of  11  min.  10.3  sec.  per  year  too  much.  In  100  yr ,  18  hr. 
37  min.  10  sec.  too  much  will  have  been  added.  Hence  every  100th 
year  is  called  a  common  year ;  but  this  would  be  dropping  too  much 
by  the  difference  between  24  hr.  and  18  hr.  37  min.  10  sec,  or  5  hr. 
23  min.  50  sec.  every  hundred  years.  By  calling  every  400th  year  a 
leap  year,  it  would  take  3840  common  years  to  equal  3840  solar  years 
and  1  day.    Hence, 

229.  Every  year  divisible  by  Jf  is  a  leap  year,  except  those 
that  have  100  and  not  JfiO  for  an  exact  divisor. 

The  names  and  order  of  the  calendar  months  are,  Janu- 
ary, 1st  mo. ;  February,  2d  mo. ;  March,  3d  mo. ;  April,  4th 
mo. ;  May,  5th  mo. ;  June,  6th  mo. ;  July,  7th  mo. ;  August, 
8th  mo. ;  September,  9th  mo. :  October,  10th  mo. ;  Novem- 
ber, 11th  mo.;  December,  12th  mo. 


TABLES. 


201 


January,  March,  May,  July,  August,  October,  and  DecemTier  have 
31  days  each. 

April,  June,  September,  and  November  have  30  days  each. 
February  has  28  days,  except  in  leap-year  when  it  has  29. 
These  numbers  may  be  remembered  by  the  aid  of  the  following 

lines  : 

Thirty  days  have  September, 

April,  June,  and  November, 

And  all  the  rest  have  thirty-one. 

Save  February,  which  alone 

Has  twenty -eight ;  and  we  assi^ 

To  this,  in  leap-year,  twenty-nine. 

MISCJELLAWIJOUS   TABLES. 
230.   Counting, 


12  things                            make  1  dozen. 

12  dozBD,  or  144  things,      "      1  gross. 

12  gross,  or  1728  things,      "      1  great  gross. 

2  things                                "      1  pair. 

6  things                                "      1  set. 

20  things                               "     1  score. 

231.  Paper. 

24  sheets                            make  1  quire. 

20  quires,  or  480  sheets,      "      1  ream. 

2  reams                                "      1  bundle. 

5  bundles                             «      1  bale. 

232.   Books. 

A  folio  book      (fo)     is  made  of  sheets  folded  in 

2  leaves. 

A  quarto            (4to)  «      "      "      "          "       " 

4 

(( 

An  octavo         (8vo)  "      "      "      "          "       " 

8 

a 

A  duodecimo  (12mo)  "      "      "      "          «       " 

12 

« 

Anl8mo                     "      "      «      "          "       " 

18 

a 

A24mo                       «      "      "      « 

24 

« 

A32mo                       "      "      ''      "          "       " 

32 

ti 

The  tervaa  fdio,  quarto,  oetaeo,  etc.,  denote  the  number  of  leaves 
into  which  a  sheet  of  paper  is  folded  in  making  books. 


20$ 


DEI^OMINATE     NUMBERS. 


333.   Copying. 

72  words  inake  1  folio,  or  sheet  of  common  law. 
90      **  "      1     "       "      *'       "  chancery. 

234.  Shoemakers^  Measure, 

Small  Sizes.— No.  1.    ^  inches. 

No.  2.    44      "       4.  J  =  44i  in. 
No.  3.    4i      "       4.  i  +  J  =:  4il  in. 
&c.,    &c.,    &c. 
Large  Sizes.— No.  1.    %^  inches. 

No.  2.     8iJ      "       +  i  =  m  in. 
No.  3.     8ii      "       +  J  +  J  =  9i    in. 
&c.,    &c.,    &c. 


t'-ii-^^^^^^ll^^m.^^S^ 


niS!ll-TJCT;IC|.lM 


i4 


^ 


235.   deduction   of  Denominate   Numhers 

consists  in  changing  their  name  without  altering  their  value. 


MONEY. 


Oi«al  iE.V 


Example  1.— How  many  dollars  in  3  E.  ?  5  E.?  10  E.  ? 
Solution.— Since  in  1  E.  there  are  10  dollars,  in  3  E.  there  are  3 
times  10  doUara  =  $30  ;  &c.,  &c.,  &c 


BEDUCTION.  203 

ritO  BZJEMS, 

1.  How  many  dollars  in  6  D.  E.  ?    20  D.  E.  ?    30  D.  E.  ? 

2.  How  many  farthings  in  3  pence  ?     2s.  ? 

3.  How  many  pence  in  3s.  ?     12s.  ?    4£  ? 

4.  How  many  s.  in  6g.?    8  cr.?     16  florins? 

5.  How  many  millimes  in  2  cL?    5  dc.  ?    10  fr.  ? 

6.  How  many  francs  in  1  Napoleon  ?    3  ?    5  ?    23  ? 

7.  How  many  c.  in  $1,  Canada  currency?  In  110  ?  In  $20? 

8.  How  many  pfennigs  in  1  mark  ?    In  10  marks  ?    In  24? 

In  the  foregoing  problems  we  have  increased  the  number  of  parts, 
diminished  the  size  of  each  part  and  changed  the  name  without  chang- 
ing the  value.  This  is  called  reducing  to  lower  denominations,  or 
Meduction  Descending ,  which  is  performed  bj  multiplication. 

Example  2.  — In  30  dollars  how  many  E.?  In  $50  ?  In 
1100? 

Solution.— Since  in  $10  there  is  1  E.,  in  $30  there  are  8  E. ;  &c.,  &c 

PROBLEMS. 

9.  In  $120  how  many  D.  E.  ?    In  $400  ?     In  $600  ? 

10.  In  14:4: far.  how  many  pence?    How  many  shillings? 

11.  In  2005.  how  many  £  ?    How  many  crowns  ? 

12.  In  126s.  how  many  g.  ?     How  many  florins  ? 

13.  In  2000?w.  how  many  centimes  ?     decimes  ?    francs  ? 

14.  How  many  Napoleons  in  20  francs  ?  In  200  ?  In  500  ? 
In  280?    In  640? 

15.  How    many  marks    in    100  pfennigs  ?     In    1000  ? 

In  2400  ?    In  2100  ?    In  10000  ? 

In  the  last  7  problems  we  have  diminished  the  number  of  parts, 
increased  the  size  of  each  part  and  changed  the  name,  without  chang- 
ing the  value  of  the  expression.  This  is  called  reducing  to  a  higher 
denomination,  or  Meduction  Ascending,  which  is  performed  by 
division. 


d04 


DENOMINATE    NUMBERS, 


MONEY. 


^IWiitton^Xorii.Hf^s  f 


236.    Simple    or    Compound     Denom^inate 
Numbers  to  Lower  Denominate  Numbers. 

Example  3.— Reduce  £140  to  farthings. 

SOLUTION. 

205. 
140 


2800s. 

nd. 

2800 
336006?. 

A:  far, 
33600 
134400 /an 

Example  4.- 


80LUTI0N. 

20s. 
5 


100s.  +  8s. 
108 


1296^.  +  5<?  =  1301</. 

Afar. 
1301* 


6204 /an +  5/ar.  =  5207 /an 


Explanation.— Since  £1  =  20«.,  £140  =  140 
times  20a.  =  2800«. 

Since  1«.  =  \M.y  2800«.  =  2800  times  VHd.  — 
336006?. 

Since  \d.  =  Afar.,  ^QOOd.  =  33600  times  Afar. 
=  134400 /ar. 

Therefore,  £140  =  134400 /an 


Reduce  £5  8s.  6d.  3  far.  to  farthings. 

Explanation.— Since  in  £1 
there  are  20».  in  £5  there  are 
5  times  20«.,  or  100«. ;  adding 
the  Ss.  given,  we  have  108«. 

Since  la.  =  12d,  108a. =108 
times  12d.,  or  1296rf. ;  which 
added  to  the  5d.  given,  makes 
1301d. 

Since  Id.  =  Afar.,  1301  d=: 
1301  times  4/an,  or  5204/ar.; 
adding  the  Sfar.  given,  we 
have  as  a  final  result  5207  far. 

Therefore,  £5  8a.  5d.  3 far. 
=  5207/ar.      Hence,  the 


108s. 


EEDUCTIOK. 


fm 


Rule. — Regard  the  number  of  the  highest  denomination 
given  as  a  multiplier^  and  the  number  of  the  next  lower 
denomination  required  to  make  1  of  this  higher  as  a  mul^ 
tiplicandf  and  to  the  product  of  these  numbers  add  the 
given  number,  if  any,  of  the  lower  denomination. 

Change  this  result  in  like  manner  to  the  next  loioer 
denomination,  and  so  proceed  till  the  required  denomination 
is  reached, 

"N amber"  in  this  rule  includes  not  only  integers,  but  also 
fractions,  both  decimal  and  common. 


JPBOB 

Reduce 

16.  325  E.  to  I. 

17.  121  D.  E.  to  $. 

18.  £8  15s.  ed.  to  pence. 

19.  £11  OS.  4:d.  3  qr.  to  qr. 

20.  £4:  M,  2  qr.  to  qr. 

21.  45  Napoleons  to/r. 

22.  150  marks  to  pfennigs. 

23.  1  crown  to  far. 


LEMS, 

Reduce 

24.  iei2  65.  3  qr.  to  qr. 

25.  i215  9s.  6d.  to  qr. 

26.  200  ct.  to  ms. 

27.  25  dc.  to  ct. 

28.  36/r.  toflfc. 

29.  3/r.  to  ms. 

30.  .075  Napoleons  to  ms. 

31.  iE7.3125  to  d. 


237.  Simple  Denom^inate  lumbers  reduced 
to  higher  Denom^inations, 

Example  5.— Reduce  134400 /ar.  to  £. 


SOLUTION. 

4  )  134400 /ar. 
12  )  33600^. 

20  )  28005. 
140  £ 


Explanation.— Since  4  far.  =  Id.,  in 
134400 /ar.  there  are  as  many  d.  as  Afar. 
are  contained  times  in  134400  far.,  or 
33600(i. ;  hence  in  134400  far.  there  are 
SSQOOd. 

Since  12(f.  =  1.^.,  33600(f.  =  2800«. 

Since  20s.  =  £1,  2800s.  =  £140.  There- 
fore, in  134400 /ar.  there  are  £140. 


DEiq^OMIKATE     NUMBERS. 


Example  6. — Keduce  5207  far.  to  a  common  denominate 
number. 


SOLUTION. 

4:)  5207  far, 
12  )  ISOld.     3  far. 
20  )_10Ss.      5d. 
5  £     85. 


Explanation.— Since  4  /ar.  =  Id, 
5207 /ar.  equal  as  many  pence  as  4/ar. 
are  contained  times  in  5207  far.,  or 
1301  pence,  with  3  remainder;  hence, 
5207 /«r.  =  1301c?.  3/ar. 

Since  \M.  =  1«„  1301d.  =  108«.  5<f. 
Since  20s.  =  £1,  108s.  =  £5  8s.    There- 


fore, 5207 /ar.  =  £5  8s.  5d  3/ar.     Hence  the  following 

EuLE. — Divide  the  given  number  hy  that  number  of  its 
denomination  ivhich  makes  one  of  the  next  higher. 

Change  the  quotient  thus  found  in  like  manner,  to  the 
next  higher  denomination,  and  so  continue  till  the  required 
denomination  is  reached. 

The  last  quotient  loith  the  several  remainders,  if  any, 
written  in  their  natural  order,  is  the  required  ansioer. 

'* Number"  in  this  rule  includes  integers,  and  decimal  and  com- 
mon fractions. 

ritOBL  EMS. 

Eeduce 


Keduce 

32.  $3250  to  E. 

33.  12420  to  D.  E. 

34.  2106d.  to  £. 

35.  6979  qr.  to  £. 

36.  3874  qr.  to  £. 

37.  900  fr.  to  Napoleons. 

38.  15000  pfennigs  to  marks. 

39.  240  far.  to  crowns. 


40.  11811  qr.  to  £. 

41.  14856  qr.  to  £. 

42.  2000  ms.  to  ct. 

43.  250  ct.  to  dc. 

44.  360  dc.  to  fr. 

45.  3000  ms.  to  fr. 

46.  1500  ms.  to  Napoleons. 

47.  1755d.  to  £. 


48.  How  many  sliillings  in  200  qrs.  ?     In  200d.? 

49.  How  many  pounds  in  2000  qr.  ?      In  2000d.  ? 
2000s.? 


In 


REDUCTION 


207 


TROY  AND  ABOTHECAHIES'  WEIGHTS. 


©rat^xfenci^eX  -^ 


"  60.  How  many  gr.  in  4  pwt.  ?     5  pwt.  ?     6  pwt.  ? 

51.  How  many  pwt.  in  4  oz.  ?     6  oz.  ?     3  lb.  ?    5  lb. 

52.  How  many  pwt.  in  960  gr.  ?    How  many  oz.  ? 

53.  How  many  lb.  in  60  oz.  ?    In  72  oz.  ?     In  160  pwt.  ? 

54.  How  many  grains  in  a  diamond  weighing  3  carats  ? 
5  carats  ?    8  carats  ? 

55.  How  many  carats  in  9.6  gr.  ?    16  gr.  ?    25.6  gr.  ? 

56.  How  much  pure  gold  in  a  ring  18  carats  fine  and 
weighing  4  pwt.  ? 


57.  How  many  gr.  in  33 

58.  How  many  3  in  4  |  ? 

59.  How  many  3  in  33? 
How  many  lb  in  12  J  ?    96  ^  ?    96  3 


53? 
125? 
243? 


60 


13?     43?    If? 
3  ft)?    51b?    111b? 
240  gr.?    365  gr.? 
-^^  -  ?     192  3  ? 


TBOY  AWn  APOTHECABTES'  WEIGHTS. 


^IWi^itten  ^xer  ci^eXi: 


^. 


Keduce 

61.  73  pwt.  to  gr. 

62.  24  lb.  to  oz. 

63.  15  pwt.  14  gr.  to  gr. 

64.  9  oz.  16  pwt.  19  gr.  to  gr. 

65.  20  lb.  3  oz.  4  pwt.  5  gr.  to  gr. 

66.  335  lb.  to  pwt. 

67.  9.325  lb.  to  gr. 


Keduce 

68.  1752  gr.  to  pwt 

69.  288  oz.  to  lb. 

70.  374  gr.  to  pwt. 

71.  4723  gr.  to  oz. 

72.  116741  gr.  to  lb. 

73.  80400  pwt.  to  lb. 

74.  53712  gr.  to  lb. 


DENOMINATE     NUMBERS. 


75.  21  §   to  3  . 

76.  51.31  §  to  3. 

77.  6.335  ft)  to  gr. 

78.  43  19  gr.  to  gr. 

79.  5  3  2  3  15  gr.  to  gr. 

80.  71  43  2b  12gr.  togr. 

81.  2  ft)  7  I  5  3  9  gr.  to  gr. 


82.  168  3  to  i. 

83.  1231.443  to  §. 

84.  36489.6  gr.  to  lb. 

85.  99  gr.  to  3. 

86.  355  gr.  to  3  • 

87.  3652  gr.  to  ft). 

88.  15189  gr.  to  §. 


AVOIRDUPOIS    WEIGHT. 


Oral  I^.vbitci^oX 


89.  How  many  drams  in  7  oz.  ?    10  oz.  ?    2  lb.  ?    10  lb.? 

Solution. — Since  in  1  oz.  there  are  16  drams,  in  7  oz.  there  are 
7  times  16  drams,  or  112  drams.  Therefore,  in  7  oz.  there  are  112 
drams. 

90.  How  many  lb.  in  1  cwt?  In  1  cental?  In  2  T.  ? 
In  5  T.? 

91.  How  many  lb.  in  512  dr.  ?    In  96  oz.  ?    In  128  oz.  ? 

92.  How  many  T.  in  100  cwt.  ?  500  cwt.  ?  4000  lb.  ? 
18000  lb.  ? 

93.  How  many  centals  in  100  lb.  ?  5030  lb.  ?  6500  lb.  ? 
7250  lb.  ? 

•    94.  How  many  lb.  in  1  long  T. ?    In  2  ?    In  3 ?    In  4? 

95.  How  many  cwt.  in  1  long  T.  ?  How  many  lb.  in  1 
long  cwt  ? 

96.  How  many  kegs  of  nails  in  100  lb.?  3001b.?  5501b.? 

97.  How  many  lb.  in  1  bu.  coal,  Pa.  ?    3  bu.  ?    10  bu.  ? 

98.  How  many  bu.  coal  in  760  T.?  2280  lb.  ?    11400  lb.  ? 

99.  How  many  lb.  in  5  tons  U.  S.  weight?  In  5  tons 
English  weight? 


REDUCTIOlf. 


209 


AVOIBDUJPOIS    WEIGHT. 


tlWi  itten  ^Xfer  ei.<es 


Eeduce 

100.  17  T.  to  oz. 

101.  213.3  oz.  to  dr. 

102.  25.375  lb.  to  oz. 

103.  163  cwt.  to  oz. 

104.  75.25  T.  to  lb. 

105.  4  long  T.  to  lb. 

106.  5  T.  3  cwt.  19  lb.  to  oz. 


Reduce 

107.  544000  oz.  to  T. 

108.  3412.8  dr.  to  oz. 

109.  406  oz.  to  lb. 

110.  260800  oz.  to  cwt 

111.  150500  lb.  to  T. 

112.  8960  lb.  to  long  T. 

113.  165104  oz.  to  T. 


MiaCELLAHEOUS. 

114.  How  many  grains  in  1  lb.  Troy  ? 

115.  How  many  grains  in  1  lb.  Av.  ? 

116.  How  many  grains  in  1  oz.  Troy?         Ans.  480  gr. 

117.  How  many  grains  in  1  oz.  Av.  ?         Ans.  437.5  gr. 

118.  How  many  centals  of  grain  in  3  long  T.  ? 

Ans,  67.2  centals. 

119.  How  many  bbl.  of  flour  in  375  cwt.  ? 

Ans.mUhU. 

120.  How  many  bu.  bituminous  coal,  Pa.,  in  1  T.  ?    How 
many  bu.  of  wheat  ?    1st  Ans.  26^^^  bu. ;  2d  Ans.  33 J-  bu. 


LONG  MEASURE. 


Oi-al  ^ 


121.  How  many  in.  in  2  ft.  ?    5  ft.  ?    6^  ft.  ?    12  ft.  ? 

122.  How  many  ft.  in  3  yd.  ?    10  yd.  ?    11^  yd. :    5^  yd? 

14 


^d 


DENOMINATE     NUMBERS. 


123.  How 

124.  How 

125.  How 

126.  How 

127.  How 

128.  How 

129.  How 

130.  In  1 

131.  How 
pace  ?    In  3 

132.  How 

133.  How 

134.  How 


many  yd.  in  2  rd.  ?     3  rd.  ?     G  rd.  ?     9  rd.  ? 
many  rd.  in  3  fur.  ?     5  fur.  ?    20  fur.  ? 
many  fur.  in  7  mi.  ?     10  mi.  ?     12|  mi.  ? 
many  mi',  in  10  lea.  ?  20  lea.  ?    60  lea.  ? 
many  statute  mi.  in  3  deg.  ?     7  deg.  ? 
many  rd.  in  33  ft.  ?    In  16J  yd.  ?    33  yd.  ? 
many  quarters  in  1  yd.  ?    3  yd.  ?     18  yd.  ? 
cable  length,  how  many  fathoms  ? 
many  in.  in  13  hands?    In  12  palms?    In  1 
cubits  ?    How  many  feet  in  5  paces  ? 
many  chains  in  3  mi.  ?     10  mi.  ? 
many  ft.  in  2  ch.  ?     7  ch.  ?    10  ch.  ?    20  ch.  ? 
many  mi.  in  160  ch.  ?    240  ch.  ?     640  ch.  ? 


LONG   MEASUJRE. 


rlWi^ittew  ^:\rGr ci.<f^.H;  t 


¥- 


Eeduce 

135.  48  ft.  to  in. 

136.  159  yd.  to  ft. 

137.  979  mi.  to  fur. 

138.  27i  lea.  to  mi. 

139.  78  rd.  9  ft.  7  in.  to  in. 

140.  87  rd.  to  in. 

141.  37  rd.  2.5  yd.  to  yd. 

142.  5  mi.  45  rd.  to  ft. 

143.  17  ch.  to  1. 

144.  720  fur.  to  ch. 

145.  3  ch.  13  1.  to  1. 

146.  3  mi.  35  ch.  65l.tol. 

147.  2  fur.  7  ch.  3  p.  to  rd. 

148.  5  mi.  25  ch.  3  p.  to  1. 


Eeduce 

149.  576  in.  to  ft. 

150.  477  ft.  to  yd. 

151.  7832  fur.  to  mi. 

152.  82  mi.  to  lea. 

153.  15559  in.  to  rd. 

154.  17226  in.  to  rd. 

155.  206  yd.  to  rd. 

156.  271421  ft.  to  mi 

157.  1700  1.  toch. 

158.  7200  ch.  to  fur. 

159.  313  1.  to  ch. 

160.  27565  1.  to  mi. 

161.  Ill  p.  to  fur. 

162.  42575  1.  to  mi 


EEDUCTION^. 


211 


SQUARE   MEASURE. 


©ral'E 


163.  How  many  sq.  in.  in  3  sq.  ft.  ?    4  sq.  ft.  ?    10  sq.  ft.  ? 

164.  How  many  sq.  ft.  in  10  sq.  yd.?     15  sq.  yd.? 

165.  How  many  P.  in  5  R.  ?   10  R.  ?  2  A.  ?  4  A.  ?  10  A.? 

166.  How  many  sq.  yd.  in  81  sq.  ft.  ?     900  sq.  ft.  ? 

167.  How  many  A.  in  8  R.  ?     24  R.  ?     320  P.  ?     640  P.  ? 

168.  How  many  sq.  yd.  in  5  A.  ?     15  A.  ?    32  A.  ? 

169.  How  many  A.  in  2  sections  ?    In  3  ?    In  5  ?    In  10  ? 

170.  How  many  sections  in  2  townships  ?     3  ?     20  ?    30? 

171.  How  many  A.  in  50  sq.  ch.  ?     150  ?     310  ?     520  ? 

172.  How  many  sections  in  1 280  A.  ?    1920  A.  ? 

Solution. — In  1  section  there  are  640  acres.    Since  Glfi  acres  a/re 
contained  in  1280  a/^res  2  times,  there  are  2  sections  in  1280  acres. 


173.  How  many  townships  in  72  sections? 
144  sections  ?     720  sections  ? 


108  sections  ? 


SQUARE  MEASURE. 


3. 


tl\¥i^ittei^  ^Xer  c  isf 


Reduce 

174.  76  sq.  ft.  to  sq.  in. 

175.  84  sq.  yd.  to  sq.  in. 

176.  170.3  P.  to  sq.  ft. 

177.  3  R.  to  sq.  ft. 

178.  7  A.  75  P.  20  sq.  yd.  to  sq.  yd. 

179.  112  A.  9  sq.  ch.  to  sq.  ch. 

180.  3  sec.  31  A.  to  A. 


Reduce 
181.  10944  sq.  in.  to  sq.ft. 
182. 108864  sq.  in.  to  sq.  yd. 

183.  46364.175  sq.  ft.  to  R 

184.  32670  sq.  ft.  to  R. 

185.  361681  sq.  yd.  to  A. 

186.  1129  sq.  ch.  to  A. 
187. 1951  A.  to  sec. 


212 


DEKOMIN^ATE     NUMBERS. 


CUBIC   MEASUBU. 


©ral^xfGPciXeX  -^ 


188.  How  many  cu.  ft.  in  3  cu.  yd.  ?     In  5  ?     In  20  ? 

189.  How  many  cu.  ft.  in  8  cd.  ft.  ?    In  2  ?    In  10  ? 

190.  How  many  cd.  ft.  in  20  cords  ?    In  30  ?    In  45  ? 

191.  How  many  perch  of  stone  in  99  cu.  ft.  ? 

192.  How  many  Eegister  tons  in  1000  cu.  ft.?    In  2000 ? 

193.  How  many  Shipping  tons  (U.  S.)  in  2800  cu.  ft.? 


CUBIC   MEASUBE. 


-^i 


tlWi^itten  ^^er  ei^e^ 


Eeduce 

194.  9cu.yd.  18cu.ft.  toft. 

195.  5  cu,Vd.  23  cu.  ft.  726 

cu.  in.  to  cu.  in. 


Eeduce 

196.  181  cd.  to  cd.  ft. 

197.  261  cu.  ft.  to  cu.  yd. 

198.  273750  cu.  in.  to  cu.  yd. 


199.  1448  cd.  ft.  to  cd. 

200.  How  many  cords  of  wood  could  be  piled  in  a  room 
16  ft.  long,  15  ft.  wide,  and  11  ft.  high?  Ans.  20|  cords. 
'     201.  How  many  cu.  yd.  in  a  cellar  30  ft.  long,  24  ft.  wide, 
and  7  feet  deep  ?  A7is,  186f  cu.  yd. 

202.  How  many  cd.  in  a  pile  of  wood  125  feet  long,  4  ft. 
wide,  and  4  ft.  high  ?  Ans.  15|  cd. 

203.  What  are  the  contents  of  a  box  42  in.  long,  27  in. 
wide,  and  2  ft.  deep?  Ans.  27216  cu.  in. 


REDUCTION. 


213 


DRY  AND  LIQUID  MEASURBS. 


©i'at3Ex:eFciXeX  '^ 


204.  How  many  qt.  in  5  pk.  ?    In  3  bii.  ?     In  10  bu.? 

205.  How  many  gi.  in  2  pt.  ?     In  3  qt.  ?    In  5  gal.  ? 

206.  How  many  ni  in  5  f  3  ?    In  2  f  ?  ?    In  30  ?  ?    In 
1  cong.  ?     In  10  cong.  ? 

207.  How  many  tablespoons  in  1  oz.  ?    In  2  wine-glasses  ? 
In  1  tea-cup  ?     In  40  tea-spoons  ? 

208.  How  many  bu.  in  128  pt.  ?     In  192  qt.  ?     In  64  pk.  ? 

209.  How  many  gal.  in  64  gi.  ?    In  24  pt.  ?     In  16  qt.  ? 


DRY  AND  LIQUID  3IEASURES. 


-.>•- 


t  Wi  itteri  ^Xfer  c  i^eX 


Reduce 

210.  8  bu.  3  pk.  to  pt. 

211.  9  bu.  1  pk.  5  qt.  to  pt. 

212.  4  bu.  3  pk.  2  qt.  1  pt.  to  pt. 

213.  3  qt.  1  pt.  3  gi.  to  gi. 

214.  40  gal.  3  qt.  2  gi.  to  gi. 

215.  5  bu.  7  qt.  1  pt.  to  pt. 

216.  5  cong.  to  f  3  . 

217.  8fl  to  f^L. 

226.  If  5  bushels  of  wheat 
many   barrels   can   be   made 
4003  qt.  ?     Out  of  8000  pt.  ? 


Eeduce 

218.  560  pt.  to  bu. 

219.  602  pt.  to  bu. 

220.  309  pt.  to  bu. 

221.  31  gi.  to  qt. 

222.  1306  gi.  to  gaL 

223.  335  pt.  to  bu. 

224.  640  f  3  to  cong. 

225.  3840  ni  to  f  S . 
make  a  barrel  of  flour,  how 

out  of    1000  gal.?     Out  of 


2U 


DEIS^OMINATE     NUMBEBS. 


CIBCVLAB  MEASURE. 


:i^1>Yi^ittei\^xrer  c  i^eX 


Eeduce 

227.  45°  28'  55"  to  seconds. 

228.  73°  58' to  seconds. 

229.  89°  20"  to  seconds. 

230.  131°  3'  29"  to  seconds. 


Reduce 

231.  163735"  to  degrees. 

232.  266280"  to  degrees. 

233.  320420"  to  degrees. 

234.  471809"  to  degrees. 


TIME  MEASURE. 


^  ©i^at  ^XGitciXeX 


235.  How  many  sec.  in  3  min.  ?    5  min.  ?    7  min.  ? 

236.  How  many  min.  in  2  hr.  ?     3  hr.  ?     5 J  hr.? 

237.  How  many  lir.  in  3  da.  ?    5  da.  ?     7  da.  ?    24  min.  ? 

238.  How  many  da.  in  3  wk.  ?     10  wk.  ?    48  hr.  ?    96  hr.  ? 

239.  How  many  hr.  in  6  da.  ?    8  da.  ?     15  da.  ?    1  wk.  ? 

TIME  3IEASUBE. 


^ 
^ 


Wi  ittcn  ^Xeivi^l^^ 


Reduce 

240.  15  hr.  25  min.  to  min. 

241.  43  min.  54  sec.  to  sec. 

242.  25  wk.  12  hr.  to  hr. 

243.  4  da.  5  hr.  6  min.  to  min. 

244.  1  cora.  yr.  to  sec. 

245.  1  solar  yr.  to  sec. 


Reduce 

246.  925  min.  to  hr. 

247.  2634  sec.  to  min. 

248.  4212  hr.  to  wk. 

249.  6066  min.  to  da. 

250.  31536000  sec.  to  da. 

251.  31556929^!^  sec.  to  da. 


RED  CrCTION 


215 


252.  How  many  days  from  June  20  to  Dec.  16  ? 

Ans.  179  da. 

253.  How  many  days  from  Jan.  16  to  Sept.  24  (1876)  ? 

Ans.  252  da. 

MISCELLANEOUS  TABLES. 


0i«al^\fei'.ci<es' 


254.  What  cost  2  gross  slate  pencils,  @  6f  b.  dozen  ? 

255.  What  cost  1  great  gross  buttons,  @  3^  a  dozen  ? 

256.  What  cost  3  pairs  vases,  @  12  apiece  ? 

257.  How  many  chairs  in  6  sets  ?     In  55  sets  ? 

258.  How  many  dozens  in  1728  ?    How  many  great  gross  ? 

259.  How  many  pairs  in  150  shoes  ?    How  many  sets  in 
96  chairs  ? 

260.  How  many  scores  in  10  gross?    In  15  gross? 

261.  How  many  quires  in  3  reams  ?    How  many  sheets  ? 

262.  How  many  bales  in  10  bundles?     How  many  reams  ? 

263.  How  many  reams  in  960  sheets  ?     In  180  quires  ? 

264.  How  many  bales  in  100  bundles  ?    In  100  reams  ? 


MISCELLANEOUS   BBOBLEMS. 


+lVVl*iUen  ^xferci^eX 


265.  A  gentleman  returning  from  Europe  brought  with 
him  320  sovereigns,  25  crown  pieces,  and  14  florins.  What 
was  their  value  in  IT.  S.  coin  ?  Ans.  $1594.50. 

266.  How  many  centals  of  wheat  can  I  purchase  for  $137, 
@  11.25  per  bu.  ?  Ans.  65.76  centals. 


316  DENOMII^ATE     NUMBERS. 

267.  Madame  de  Shiel  presented  to  a  friend  in  Florida, 
125  Napoleons  and  17  francs.  What  was  the  value  of  the 
present  in  U.  S.  money?  Ans.  $485,781. 

268.  I  received  in  change  on  a  Canadian  Railroad,  13  dol- 
lars, 5  50-cent  pieces,  and  3  20-cent  pieces.  What  is  their 
value  in  the  U.  S.  ? 

269.  An  estate  in  Germany  yields  me  an  annual  income 
of  3000  marks.    How  much  is  that  in  U.  S.  money  ? 

270.  A  druggist  bought  576  lb.  of  licorice  at  30)^  per  lb. 
Av.,  and  sold  at  the  same  price  per  pound  Apoth.  weight. 
How  much  did  he  gain  ?  Ans.  $37.20. 

271.  What  cost.  250  bu.  bituminous  coal,  Pa.,  at  $2.60 
per  ton?  A7is.  $24.70. 

272.  What  cost  35860  lb.  salt  at  K  Y.  salt  works,  at  $1.75 
per  bbl.  ?  Ans.  $224.12. 

273.  A  jeweler  purchased  13  gold  watch-cases,  each  weigh- 
ing 16  pwt.,  18  carats  fine,  at  $15  per  oz.  for  the  gold,  the 
alloy  to  count  for  nothing.     What  did  the  cases  cost  him  ? 

274.  A' confectioner  sold  5  casks  of  raisins  at  15  cents  per 
lb.     How  much  did  he  receive  for  them  ? 

275.  A  gentleman  expecting  to  travel  in  Great  Britain 
exchanged  $1600  for  English  money.  How  many  pounds, 
&c.,  did  he  receive  ?  Ans.  £328  15s.  6.8d. 

276.  When  he  arrived  in  London,  he  concluded  lo  go  to 
France,  and  then  exchanged  £100  into  French  money. 
What  did  he  receive  ?  Ans.  2521  fr.  50  ct. 

277.  What  cost  15  bu.  and  3  gal.  of  wheat  at  $.02  a  pint  ? 
At  $.005  a  gill  ? 

278.  A  man  paid  $6.40  for  a  ton  of  sand  ;  how  much  did 
he  pay  for  each  pound  ?    For  each  ounce  ? 

279.  A  man  paid  $2.50  for  a  barrel  of  salt;  how  much 
did  he  pay  per  ounce  ? 


EEDUCTIOK.  217 

238.  Denominate  Fractions  to  Higher  or 
Lower  Terms* 

A  Denom^inate  Fraction  is  one  which  expresses 
one  or  more  equal  parts  of  a  denominate  unit. 


0i"^at  ^t&fciXgX 


Example  1. — Reduce  £^  to  a  fraction  of  a  shilling. 
Solution.— Since  in  £1  there  are  20s.,  in  £^  there  are  yV  o^  ^Os. 
=  |s.,  or  l^s. 

FBOBT.EMS, 

1.  Eeduce  ^V  ^^  ^  shilling  to  the  fraction  of  a  penny. 

2.  Reduce  :^  lb.  Troy  to  the  fraction  of  an  ounce. 

3.  Reduce  ^j  lb.  Apoth.  to  the  fraction  of  an  ounce. 

4.  Reduce  .  07  bu.  to  the  decimal  of  a  peck. 
6.  Reduce  .253  to  the  fraction  of  a  3. 

6.  Reduce  .025  mi.  to  the  fraction  of  a  furlong. 

7.  Reduce  .125  yd.  to  the  fraction  of  a  foot. 
Example  2. — Reduce  1  Js.  to  the  fraction  of  a  £. 

Solution. — Since  20s.  =  £1,  there  are  as  many  pounds  in  1\,  or 
|s.  as  20  shillings  are  contained  times  in  f  shillings,  or  /^^  =  £^. 
Therefore,  l^s.  =  £xV. 

8.  Reduce  f  of  a  penny  to  the  fraction  of  a  shilling. 

9.  Reduce  ^^  oz.  Troy  to  the  fraction  of  a  pound. 

10.  Reduce  2^  oz.  Apoth.  to  the  fraction  of  a  pound. 

11.  Reduce  .28  pk.  to  the  fraction  of  a  bushel. 

12.  Reduce  .75  3  to  the  fraction  of  a  3  • 

13.  Reduce  .2  fur.  to  the  fraction  of  a  mile. 

14.  Reduce  .375  ft.  to  the  fraction  of  a  yard. 

It  will  be  observed  that,  as  in  integers,  Hediiction  Descending 
is  performed  by  Multiplication,  and  Reduction  Ascending  by 
Division. 


^Is 


DENOMINATE     NUMBEBS 


^i^ittew  ^.^ert  i;^eX  i 


Rules.— Same  as  for  Integers  (237). 


Reduction  Descending, 
Ex.  3. — Reduce  ^yifcr  to  d. 


SOLUTION. 
31 


Ad. 


Reduction  Ascending. 
Ex.  4.— Reduce  ^  d  to  £. 

SOLUTION. 


PnOBL  EMS. 


15. 

16. 
17. 
18. 
19. 
20. 
21. 
22. 
23. 
24. 
25. 
26. 
27. 
28. 
29. 
30. 
31. 
32. 


T^^  g.    to  the  fraction  of  a  d. 

tH  florin  " 
rfuPwt.    " 


6i 


te 


IwT. 


«  lb. 
"  ft. 


in. 
«  1. 


tAtt  sq.  yd. 

1         A  a 

TTT8-  ^* 

.025  R  to  P. 
.015  R.  to  sq.  1. 
.027  cd.  ft.  to  cu.  ft. 
.037  cd.  to  cu.  ft. 
.625  bu.  to  pk. 
.832567  pk.  to  pt. 
.0025  com.  yr.  to  da. 
.00125°  to  seconds. 


sq.  m. 
"  sq.  yd 


33.  :^Q  d.     to  the  fraction  of  a  g. 

34.  U  d.        " 

35.  *  gr.      " 

36.  m  3-    " 


"  flor. 
«  pwt. 


37.  iM  lb. 


38.  %V-ft.    "    " 

39.  11^  in.      «    « 

40.MJL      "    " 
41.  ^\  sq.  in.      " 

42.^^sq.yd.   « 
43. 1  P.  to  R. 

44.  375  sq.  1.  to  R. 

45.  .432  cu.  ft.  to  cd.  ft. 

46.  4.736  cu.  ft.  to  cd. 

47.  2.5  pk.  to  bu. 

48. 13.321072  pt.  to  pk. 
49.  .9125  da.  to  com.  yr. 
60.  4.5"  to  degrees. 


T. 

fur. 
mi. 
fur. 
sq.  yd. 
A. 


BEDUCTIOK. 


219 


239.  Simple  Denominate  Fractions  to  Sim- 
ple or  Compound  Denominate  Numbers. 

Example  1. — What  is  the  value  of  £|  ? 

Explanation. — Since  £1 

SOLUTION. 
(£1  =r  ^^S.)  Xi=  J-PS.  =  1HS. 

(ls.  =  ifd.)xi  =  Yd.  =  lid. 
(Id.  =  t  far.)  X  i  =  4  far.  =  1^  far. 
Result,  lis.  Id.  1^  far. 


=20s.,£f=f  of  20s.,orllis. 

Since  ls.=  13d.,  ^s.=  \  of 
12d.,  or  lid. 

Since  Id.  =  4  far.,  id.=  ^ 
of  4  far.,  or  1^  far. 

Therefore,  £|  =  lls.  Id. 
11  far. 


Example  2.— What  is  the  value  of  .5375  rd.  ? 

SOLUTION. 

5.5  yd. 
.53?5 


26875 
26875 


2.95625  yd. 

3  ft. 
.95625 

2.86875  ft. 

12  in. 
.86875 


Explanation.— Since  1  rd.  = 
5^  yd.,  .5375  rd.=.5375  x  5.5  yd.,  or 
2.95625  yd. 

Since  1  yd.=  3  ft.,  .95625  yd.= 
.95625  X  3  ft.,  or  2.86875  ft. 

Since  1  ft.  =  12  in.,  .86875  ft.= 
.86875  X  12  in.,  or  10.425  in. 

Therefore,  .5375  rd.=  2  yd.  2  ft. 
10.425  in. 


10.42500  in. 


Result,  2  yd.  2  ft.  10.425  in. 

KuLE. —  Use  the  given  fraction  or  decimal  as  a  Multiplier, 
and  the  number  of  the  next  lower  denomination  that  it  re- 
quires to  make  1  of  the  given  denomination  as  a  Multipli- 
cand. 

In  the  same  manner  change  the  fractional  or  decimal  part 


^20 


DEl^OMINATE     KUMBERS. 


of  the  product  thus  obtained;  and  so  proceed  till  the  required 
denomination  is  reached. 

The  integers  and  the  last  result  arranged  in  their  natural 
order,  is  the  answer  required. 


PltOBLEMS. 

What  is  the  valne 


1.  Of  £|?  Ans.  12s.  6d. 

2.  Of  I  mi.? 

^W5. 4  fur.  17  rd.  12  ft.  10  in. 

3.  Of  f  ch.?^W5.52ft.  9|in. 

4.  Of  I  A.?  ^ws.  2  R.  8  P.  242  ft. 

5.  Of  4  cd.?  Ans.  91  ft.  740^  in. 


6.  Of  .375  1b.  Troy? 

7.  Of  .3275ft).Apoth.? 

8.  Of  .751  gal.  ? 

9.  Of  .3756  bu.  ? 

10.  Of  .757circum.? 

11.  Of  .075  sign? 


12.  What  is  the  value  of  ^  of  a  bushel  ? 

Ans.  1  pk.  3  qt.  -^  pt 

13.  What  is  the  value  of  f  of  a  cwt.  ? 

Ans.  88  lb.  14  oz.  ^  dr. 

14.  What  is  the  value  of  .751  yr.  (365^  da.)  ? 

Ans,  274  da.  7  hr.  15  min.  57.6  sec. 

15.  What  is  the  value  of  .567  of  a  sign  ?    Ans.  17°  36". 

16.  What  is  the  value  of  .3752  reams  paper  ? 

Ans.  7  quires  12.096  sheets. 

240.  A  Simple  or  Compound  Denominate 
Number  to  a  Sim^ple  Denominate  Ft^action 


©Fat'^xfoicis^hX 


Example  1.— What  part  of  a  lb.  Troy  are  3  oz.  ?    5  oz.  ? 

^  Solution.— Since  in  12  oz.  there  is  1  lb.,  in  3  oz.  there  are  as  many 
lb.  as  12  oz.  are  contained  times  in  3  oz.,  or  y\,  or  \  lb.  Therefore,  in 
8  oz.  Troy  there  is  \  lb. 


EEDUCTION. 


221 


PB  OBIjEMS. 

1.  What  part  of  1  bu.  are  2  pk.  ?     3  pk.  ?     12  qt.  ? 

2.  What  part  of  1  fur.  are  5  rd.  ?     16  rd.  ?    25  rd.  ? 

3.  What  decimal  of  a  £  equals  5s.  ?    8s.  ?    15s.  ?     16s.  ? 

4.  What  decimal  of  a  da.  equals  6  lir.  ?     9  hr.  ?     15  hr.  ? 

5.  What  decimal  of  a  gal.  equals  3  pt.  ?    5  pt.  ?     7  pt.  ? 

6.  What  decimal  of  a  cord  equals  32  ft.  ?    40  ft.  ?    96  ft.  ? 


h  Written  ^7&pci^eXt 


Example  2. — What  part  of  a  bushel  equals  3  pk.  2  qt.  ? 


1st  solution. 
8 )  2  qt. 

4 )  3.25  pk. 

.8125  bu. 

2d  solution. 
2qt.:=fpk.=ipk. 

3pk.  +  ipk.=:3ipk.=J^pk. 
J^pk.=ifbu. 


1st  Explanation— Since  8 
qt. = 1  pk. ,  2  qt. = .25  pk.  3  pk. 
+  .25pk.=3.25pk. 

Since  4  pk.=l  bu  ,  3.25  pk.= 
^^,  or  .8125  bu.  Therefore, 
3  pk.  2  qt.=.8125  bu. 

2d  Explanation.— Since  8 
qt.  =  l  pk.,  2  qt.=f,  or  \  pk. 
3  pk.  +  ^  pk.=3],  or  Y"  pk. 

Since  4  pk.=:l  bu.,  ^  pk.= 
(i38-,.4)  =  lfbu. 


Rule. — If  the  numher  be  simple,  proceed  according  to 
(237). 

If  the  number  be  compound,  divide  the  lowest  given  de- 
nomination by  the  number  of  that  denomination  required  to 
make  one  of  the  next  higher.  To  this  result  add  the  units, 
if  any, of  this  higher  denomination. 

Change  the  result  thus  obtained  in  the  same  way ;  and  so 
proceed  till  the  required  denomination  is  reached, 

Tlie  last  result  will  be  the  answer  required. 


?^  DENOMII^ATE     NUMBERS. 

1*11  OBLEMS. 

7.  Wliat  part  of  a  T.  are  7  cwt  16  lb.  ?       Ans.  ^JJ  T. 

8.  What  part  of  a  lea.  are  2  mi.  3  fur.  6  rd.  ? 

Ans.  HJ  lea. 

9.  Express  in  decimals  the  part  20  sq.  yd.  72  sq.  in.  are  of 
a  rood.  Ans.  .0165748+  R. 

10.  Change  5  pwt.  15  gr.  to  the  decimal  of  a  lb. 

Ans.  .0234375  lb. 

11.  Eeduce  5  hr.  48  min.  46.05  sec.  to  the  decimal  of  a  da. 

Ans.  .242199+  da. 

341.  To  find  the  Part  that  one  Denominate 
Number  is  of  Another. 

Example.  — What  part  of  £1  are  7s.  6d.  ? 

SOLUTION  Explanation. — Since  things  compared 

7     fi  1  QOrl         ^       must  be  of  the  same  kind,  78.  Gd.  must  be 

7S^^.  _  _yua.  __  o      ^^^^  ^^^^  ^^  ^^  ^^^^^  g^^  ^^  ^^  2^^^ 

£1  240d.        8       that  is  ^4%  or  f. 

Rule  I. — Reduce  both  numbers  to  the  same  denomination  ; 
then  write  that  number  tvhich  is  to  be  measured  for  a  nu- 
merator, and  the  number  to  be  used  as  a  measure  for  a 
denominator, 

PROBTjEMS. 

1.  What  part  of  $5  are  25c.  ? 

2.  What  part  of  2  mi.  are  280  rd.  ? 

3.  What  part  of  4  A.  are  1  A.  3  R.  25  P.  ? 

4.  What  part  of  1  cd.  is  5  ft.  by  4  by  3? 

5.  What  part  of  9  gal.  are  3  gal.  3  qt  ? 

6.  What  part  of  7  bu.  are  5  bu.  1  pk.  ? 

7.  What  part  of  1  lb.  Av.  is  1  lb.  Apoth.  ? 

8.  What  part  of  1  da.  are  6  hr.  ? 

9.  What  part  of  $1  are  6Jc.? 

10.  What  part  of  1  ?  Apoth.  is  1  oz.  Av.  ? 

11.  What  part  of  1  score  is  1  doz.? 


Ans. 

A- 

Ans. 

tV 

Ans. 

tVs- 

Ans. 

«• 

Ans. 

A- 

Ans.  }. 

Ans. 

m- 

Ans.  \. 

Ans. 

A- 

Ans. 

m- 

Ans.  |. 

SECTION    IV, 


243. 


OpaigE^vfertciX'oX  "^ 


Example. — What  is  the  sum  of  £3  7s.  and  £2  19s.  ? 

Solution.— 7s.  and  19s.  =  26s.  =  £1  and  6s.  £3  and  £2  and  the 
£1  obtained  by  adding  the  shillings,  equal  £6.  Therefore  the  sum  of 
£8  78.  and  £2  19s.  is  £6  6s. 


PHOBL  EMS. 

1.  Add  together  3s.  4d.  and  2s.  6d. 

2.  Find  the  sum  5  |  3  3  and  7  3  • 

3.  How  much  are  10  ft.  5  in.  and  3  ft.  9  in.  ? 

4.  Find  the  sum  of  3  mi.  4  fur.  and  3  fur.  20  rd. 

5.  How  much  is  J  ib.  Troy  and  J  oz.  Troy  ? 

6.  Find  the  sum  of  3^  cd.  and  4  cd.  ft. 

7.  Find  the  sum  of  J  da.  and  J  da.  9  hr.  13  min. 

8.  Find  the  sum  of  J  gross  +  3  score  -f  1  set. 

9.  J.  Logan  bought  f  of  an  acre  from  one  man,  and  f  of  a 
rood  from  another.     How  much  did  he  buy  ? 

10.  Sold  f  of  a  ton  of  iron  to  one  man  and  f  of  a  cwt.  to 
another.    How  many  pounds  were  sold  ? 


224 


DENOMIITATE     NUMBERS. 


sWi'itten  lEj&rci^eXt 


bu. 

SOLUTION, 
pk.        qt. 

pt. 

15 

2 

3 

0 

23 

3 

0 

1 

1 

4 

1 

3 

2 

0 

1 

43 


0 


Example  1. — Find  the  sum  of  15  bu.  2  pk.  3  qt. ;  23  bu. 
3  pk.  1  pt. ;  1  pk.  4  qt.  1  pt. ;  and  3  bu.  2  pk.  1  pt. 

Explanation. — Since  only  like  concrete 
numbers  can  be  added  ( 62  ),  for  conveni- 
ence we  write  the  numbers  representing 
like  denominations  in  a  separate  column, 
being  careful  to  arrange  the  orders  of  units 
as  in  Addition  of  Simple  Numbers. 

Next  find  the  sum  of  the  numbers  in  the 
column  of  the  lowest  denomination  ;  this 
is  3  pt.  =  1  qt.  and  1  pt.     Write  1  under 
the  column  of  pt.  and  reserve  the  1  qt.  to 
be  added  to  the  column  of  qt. 

The  sum  of  the  column  of  qt.  =  7  qt.,  which  added  to  the  1  qt. 
reserved  =  8  qt.  =  1  pk.  Write  0  under  the  column  of  qt.  and  re- 
serve the  1  pk.  to  be  added  to  the  column  of  pk. 

The  sum  of  the  column  of  pk.  =  8  pk.,  which  added  to  the  1  pk. 
reserved  =  9  pk.  =  2  bu.  1  pk.  Write  1  pk.  under  column  of  pk. 
and  add  the  1  bu.  to  the  column  of  bu.,  the  sum  of  which  is  42  bu. 
42  bu.  +  1  bu.  =  43  bu. 

Example  2.— Find  the  sum  of  3.25  pk.  and  .953125  bu. 

1st  SOLUTION. 

3.25  pk.  =  .8125      bu. 
.953125  bu. 


1.765625  bu. 
4 


3.062500  pk. 
8 


1st  Explanation. — Reduce  the  3.25 
pk.  to  the  decimal  of  a  bu.  ( 240 ). 
Then  add  the  decimals  and  reduce 
their  sum  by  (239). 


.5000 
2 


qt. 
1.0000     pt. 


Result,  1  bu.  3  pk.  1  pt. 


ADDITIOlf.  225 


2d  solution. 

bu.  pk.  qt.  pt.  2d  Explanation.— Reduce 

each  decimal  to  a  denominate 
or  compound  denominate  num- 
ber, and  then  add  as  in  Ex,  1. 


3.25  pk.  =         3     2     0 

.953125  bu.  =         3     6     1 


Kesult,        13    0    1 


Example  3. — Find  the  sum  of  f  wk.  and  |  da. 

1st  Explana- 

IST  solution.  tion.— Reduce  f 

I  wk.  =  J^  da.  =  ^  da.  wk.  to  da.    Then 

I  da.  —   I  da.  *       add  V^  da.  and  f 

ij^  da.  -f  4  da.  =  ^  da.  =  5  da.  5  hr.  20  min.  ^a.  as  in  (181), 

and  reduce  their 
sum  to  a  compound  denominate  number  (239). 


2d  SOLUTION. 

da.  hr.  min.  2d   Explanation.  —  Reduce  both  | 

2  ^-^^  r=  4     16        0  "^k.  and  |  da.  to  compound  denominate 

^    ,  -,  f.  ««  numbers,  and  then  find  their  sum  as  in 

i  <ia.  =  13  20  j,^  ^ 

5       5     20 

KuLE  I. —  Write  the  numbers  so  that  those  representing 
like  denominations  shall  stand  in  the  same  column^  arrang- 
ing the  orders  of  units  as  in  addition  of  simple  numbers. 

Find  the  sum  of  the  numbers  of  the  lowest  denomination, 
and  reduce  this  sum  to  the  next  higher  denomination. 

Write  the  remainder,  if  any,  under  the  column  added  and 
add  the  quotient  to  the  column  of  the  next  higher  denom- 
ination. 

Do  this  with  all  the  denominations  to  the  highest,  of  which 
write  the  whole  sum, 

15 


226  DENOMINATE     NUMBERS. 

n. — Reduce  decimal  or  common  fractions  to  the  same  de- 
nomination ;  find  their  sum  and  reduce  it  to  a  denominate 
number.     Or, 

Reduce  decimals  or  fractions  to  denominate  numbers;  then 
find  their  sum. 


PBOBZEMS, 

(1.) 

(2.) 

(3.) 

£       s.       d.    far. 

£ 

8. 

d.  far. 

gal. 

qt.    pt. 

gi. 

12     17     10     3* 

25 

8 

6 

21 

3     1 

3 

9     14       9     1 

18 

9     3 

4 

2    0 

2 

14       4     2 

38 

15 

4     2 

7 

1     1 

3 

23      7       0    2 

65 

2 

8     1 

34 
(5.) 

0     0 

0 

(4.) 

ml.  fiir.  rd.  yd. 

ft. 

in. 

mi. 

far. 

rd. 

ft 

in. 

9     6     16     4 

2 

0 

12 

6 

33 

12 

7 

5     7     25     2 

2 

6 

10 

5 

18 

4 

8 

4     5     12     1 

0 

10 

16 

5 

40 

5 

6 

20    3    14    2(1)     3      4  40    3    12      5(J)    9 

(0  =  1     C a)  =  6 

20    3    14    3  0    10  40    3    13      6        3 


(6.)  (7.)  (8.) 

ml   fur.  ch.   p.  1. 

13     7     6     2  19 

25     4    6     3  14 

38    0    4    1  16 

77    4    6    3  24 

63      5 


mi. 

fur. 

ch. 

p. 

1. 

cd.  cd.ft. 

cu.ft 

19 

6 

7 

3 

21 

21      7 

15 

12 

6 

8 

2 

15 

26      3 

13 

6 

3 

9 

1 

4 

13      2 
1      6 

7 

39 

0 

5 

3 

15 

15 

ADDITION.  227 

(9.)  (10.) 

A.     R.      P.     sq.  ft.         sq.  in.  A.        P.     sq.  yd.      sq.  ft.  sq.  in. 


13 

2  25  243 

96 

15  120 

20 

5 

72 

19 

3  30   32 

50 

20   45 

5 

7 

84 

20 

2  13    1 

0  29    4(f) 

40 
42 

30   15 

5 

2 

0 

54 

66   21 

(f) 

6 

12 

(f)  = 

.108 

(i) 

=  6 

108 

54 

0  29    5 
(11.) 

6 

66      21 
(12.) 

1 

3 
(13.) 

120 

lb. 

oz.  pwt.  gr. 

lb. 

oz.  pwt.  gr. 

cd. 

cu.  ft. 

cu.  in. 

25 

6  13  20 

19 

9  15   0 

249 

73 

1024 

8 

9  17  19 

29 

12  17  23 

87 

95 

1652 

32 

10  19  16 
3  11   7 

10  18  22 

9 
347 

120 
34 

1533 

67 

50 

9  11  21 

753 

(14.) 

(15.) 

(16.) 

lb. 

1   3  3  gr. 

lb. 

!   3  3  gr. 

en.  yd. 

cu.  ft. 

cu.  in. 

6 

11  7  2  19 

5 

8  6  2  17 

146 

25 

1516 

7 

5  4  0  12 

2 

9  7  2  16 

278 

20 

934 

8 

9  6  1  15 

10  6  1  16 

4352 

4 

315 

23 

3  2  2   6 

9 

5  5  19 

4777 

23 

1037 

17.  Find  the  sum  of  15  T.  13  cwt.  23  lb.  14  oz.  13  dr.; 
7  T.  12  cwt.  19  lb. ;  17  cwt.  15  lb.  15  oz. ;  and  24  lb.  12  oz. 

Ans.  24  T.  2  cwt.  83  lb.  9  oz.  13  dr. 

18.  Find  the  sum  in  long  ton  weight  of  6  T.  14  cwt.  2  qr. 
26  lb.  12  oz.  10  dr. ;  9  T.  18  cwt.  3  qr.  27  lb.  8  oz.  VZ  dr. ; 
and  2  T.  3  cwt.  1  qr.  16  lb.  10  oz.  2  dr. 

Ans.  18  T.  17  cwt.  14  lb.  15  oz.  8  dr. 


228  DENOMINATE     NUMBEES. 

19.  The  contents  of  4  barrels  measured  37  gal.  3  qt.  1  pt. 
3  gi. ;  39  gal.  2  qt.  1  gi. ;  40  gal.  2  qt.  1  pt.  2  gi.;  and 
41  gal.  2  qt.  1  pt.  1  gi.    Find  the  whole  contents. 

Ans.  159  gal.  3  qt.  3  gi. 

20.  A  druggist  sold  laudanum  as  follows :    10  f  §  3  f  3  ; 

2  0.  4fl  7f3;  12f!  5f3  ;  6f3  50  fr^;  7f3  45^1;  and 

3  f  3  57  ^U.    What  was  the  amount  sold  ? 

Ans.  3  0.  14  f§  lf3  32ni. 

21.  Four  farmers  raised  wheat  one  year  as  follows  :  A  raised 
75  hu.  3  pk.  5  qt.  1  pt;  B,  225  bu.  1  pk.  6  qt;  C,  135  bu. 
2  pk.  4  qt  1  pt ;  and  D,  96  bu.  3  pk.  7  qt  1  pt  How  much 
did  they  all  raise?  Ans.  533  bu.  3  pk.  7  qt.  1  pt. 

22.  How  much  time  is  there  from  2  hr.  51  min.  35  sec. 
before  12  m.  to  35  min.  45  sec.  past  2  o'clock  p.  m.  ? 

Ans.  5  hr.  27  min.  20  sec. 

23.  Find  the  sum  of  13  wk.  4  da.  5  hr.  30  min.  37  sec; 
2  wk.  5  da.  22  hr.  45  min.  52  sec. ;  6  wk.  3  da.  20  hr.  50  min. 
28  sec. ;  and  6  da.  15  hr.  33  min. 

Ans.  23  wk.  6  da.  16  hr.  39  min.  57  sec. 

24.  A  has  16  quires  8  sheets  of  paper;  B,  18  quires 
15  sheets;  C,  1  ream  10  quires  13  sheets;  and  D,  3  reams 
17  quires  9  sheets.     How  much  paper  have  all  four  ? 

Ans.  7  reams  2  quires  21  sheets. 

25.  A  has  2  great  gross  5  gross  10  doz.  and  5  pencils ; 
B,  2  great  gross  10  gross  10  doz.  and  8 ;  0,  3  great  gross 
2  gross  2  doz.  and  3 ;  and  D,  7  gross  4  doz.  and  4.  How 
many  have  all  ?     Ans.  9  great  gross  2  gross  3  doz.  and  8. 

26.  The  latitude  of  Pittsburgh  is  40°  27'  36"  North,  and 
the  Cape  of  Good  Hope  is  in  33°  56'  3"  South  latitude. 
Through  how  many  degrees  of  latitude  must  one  travel  to 
go  from  Pittsburgh  to  the  Equator  ?    From  the  Equator  to 


ADDITIOi^.  229 

the  Cape  of  Good  Hope  ?    From  Pittsburgh  to  the  Cape  of 
Good  Hope  ?  Ans.  to  last,  74°  23'  39". 

27.  Chicago  is  10°  33'  40.35"  West  longitude  from  Wash- 
ington City;  and  New  York  is  3°  3'  59.2"  East  longitude 
from  Washington.  Through  how  many  degrees  of  longitude 
must  a  person  travel  in  going  from  Chicago  to  Washington  ? 
In  going  from  Chicago  to  New  York  ? 

Ans.  to  last,  13°  37'  39.55". 

28.  What  is  the  sum  of  £.628125,  £.284375,  and  .15s.? 

A71S.  18s.  4d.  3.2  far. 

29.  Find  the  sum  of  .703125  mi.,  .965625  mi.,  and  .025  fur. 

Ans.  1  mi.  5  fur.  15  rd. 

30.  Find  the  sum  of  .282  T.,  .96  cwt.,  and  .325  lb. 

Ans.  6  cwt.  60  lb.  5.2  oz. 

31.  Find  the  sum  of  .815625  lb.  Troy,  .7  lb.  Troy,  .16  oz. 
Troy,  and  13.1  gr.  Ans.  1  lb.  6  oz.  7  pwt.  11.9  gr. 

32.  Find  the  sum  of  ^s.  and  £f.  Ans,  16s.  6d. 

33.  Add  I  rd.  to  |  mi.  Ans.  214  rd. 

34.  Add  I  p.  to  If  ch.  Ans.  2  ch. 

35.  Add  5J  acres  to  3|  roods.         Ans.  6  A.  3  R.  12  P. 

36.  What  is  the  sum  of  4|  cords  and  6f  cord  feet? 

Ans.  5  cd.  5  cd.  ft.  12  cu.  ft. 

37.  To  f  lb.  Troy  add  f  oz.     Ans.  9  oz.  9  pwt.  4ff  gr. 

38.  What  is  the  sum  of  f  of  a  ton  and  |  cwt.  ? 

Ans.  13  cwt.  88  lb.  14  oz.  3|  dr. 

39.  What  is  the  sum  of  |  of  a  gallon  and  ^  of  a  gallon  ? 

Ans.  1  gal.  2  qt.  1-J-  gi. 

40.  What  is  the  sum  of  |-  of  a  circumference,  f  of  a 
degree,  and  -^  of  a  minute  ?  Ans.  200°  38'  5". 

41.  What  is  the  sum  of  |  of  a  mile,  f  of  a  furlong  and  -| 
of  a  rod  ?  Ans.  5  fur.  9  rd.  3  yd.  2  ft.  11  in. 


m  UBJT«i  A€'T'IO;Bi 


243. 


©paltE:^: 


Example. — ^Wbat  is  the  difference  between  3  gal.  2  qt., 

and  1  gal.  2  qt.  ? 

Solution.— 2  qt.  from  2  qt.  leave  nothing  ;  1  gal.  from  3  gal.  leaves 
2  gal.  Therefore,  the  difference  between  3  gal.  2  qt,,  and  1  gal.  2  qt., 
is  2  gal. 

1.  Find  the  difference  between  10  bu.  3  pk.,  and  5  bii. 
2  pk. ;  6  bu.  1  pk.  ;  7  bu.  3  pk. ;  8  bu.  2  pk.  ;  9  bu.  3  pk. 

2.  What  is  left  after  taking  from  .5  bu.  .35  bu.  ?  .22  bu.? 
.16  bu.  ?     .125  bu.  ?     .64  pk.  ?     .72  pk.  ?    f  pk.  ? 

3.  Take  from  f  mi.  |  mi. ;     -^  fur.  ;    |  fur. ;    -^^  fur. 

4.  How  much  is  -f  ch.  less  |  rd.  ?  less  25  1.  ?  less  20f  ft.  ? 
less  .4  ch.  ?    less  .25  ft.  ?    less  10.5  1.  ? 

5.  A  merchant  bought  a  hogshead  of  molasses  containing 
54  gallons,  and  sold  from  it  13  gallons  and  2  quarts  ;  how 
much  remained  ? 

6.  A  man  owning  40  acres  2  roods  and  20  perches  of 
land,  sold  9^  acres  and  .5  of  a  rood  ;  how  much  remained  ? 

7.  4^  tons  +  10.5  cwt.  -  1.25  tons  =  ? 

2») 


SUBTliACTIOK. 


231 


a 


-♦-•- 


t  Wi'itten  ^^fercisesT? 


Example  1.— From  19  cwt.  47  lb.  5  oz.,  take  12  cwt. 
96  lb.  13  oz. 


SOLUTION, 
cwt.  lb,  OZ. 

19       47        5 
12       9G       13 


6      50 


8 


Explanation.  —  We  first,  as   a  matter 

of  convenience,  write  tlie  numbers  of    tlio 

subtrahend  under  tliose  of  the  minuend,  so 

that  the  denominations   of  the  same  name 

stand  in  the  same  column,  with  units  under 

units,  tens  under  tens,  &c.     For  convenience, 

we  commence  at  the  lowest  denomination, 

and  say  mentally,  13  oz.  from  5  oz.,  impossible ;  but  adding  1  lb.  = 

16  oz.  to  5  oz.  of  the  minuend,  we  have  21  oz.,  from  which  subtracting 

13  oz.  there  remain  8  oz.  ;  and  we  write  8  under  the  oz.  column. 

Since  16  oz.  (=  1  lb.)  have  been  added  to  the  minuend,  we  must 
add  1  lb.  to  the  subtrahend  (70). 

97  lb.  from  47  lb.,  impossible  ;  to  47  lb.  we  add  100  lb.  (=  1  cwt.) 
making  147  lb.  ;  97  lb.  from  147  lb.  leave  50  lb. ;  and  we  write  50 
imder  the  lb.  column. 

We  then  add  1  cwt.  (=  100  lb.)  to  13  cwt.,  making  13  cwt. ;  13  cwt. 
from  19  cwt.  leave  6  cwt.     Result,  6  cwt.  50  lb.  8  oz. 
The  preceding  explanation  may  be  illustrated  as  follows: 

CWT.  LB.  oz. 

19  =19  cwt.  .  47  +  (100  lb.)=147  lb.  .      5 +  (16  oz.)=21  oz. 

12  +  (lcwt.):^13  cwt.  .  90  +  (    1  Ib.')^  97  lb.  .  13 =13  oz. 

6  (6  cwt.)  50  (50  1b.)       8  (8oz.) 

It  will  readily  be  observed  that  100  lb.  16  oz.  (=  101  lb.)  have  been 
added  to  the  minuend  ;  and  1  cwt.  1  lb.  (=  101  lb.)  have  been  added 
to  the  subtrahend.  The  difference,  therefore,  remains  the  same  (76). 
The  dark  letters  show  the  original  example  ;  the  quantities  in  paren- 
theses the  addends. 

Example  2.  —From  -J  bu.  take  ^  pk. 

1st  solution. 
i  bu.— ^  pk.=J  pk.-^  pk.=3-^  pk.=3  pk.  2  qi  1^^  pt 


DENOMINATE     NUMBERS. 

Explanation. — We  reduce  the  de- 

2d  solution.  nominate  fractions  to  the  same  denom- 

pk.    qt.  pt.  ination ;  then  subtract,  and  reduce  the 

-J  bu.  =3     4     0  remainder  to  a  simple  or  compound  de- 

■A-  pk.  =         1      044  nominate  number.    Or,  we  reduce  the 

■—     7     ~~  denominate  fractions  to  simple  or  com- 

"     '^       TT  pound  denominate  numbers  and  then 

subtract  as  in  the  preceding  example. 

Example  3.— From  129.716  cords  take  15.048  cd.  ft. 

1st  solution. 

129.716  cd.  Explanation.— We  re- 

15.048  cd.  ft.  =       1.881  cd.  duce    both    numbers    to 

cords,  then  subtract  and 
reduce  the  remainder  to 
cords  and  cu.  ft.     Or, 

We  may  reduce  both 
cords  and  cord  feet  to 
simple,  or  compound  de- 
nominate numbers,  and 
then  subtract  as  in  Ex.  1. 

127     106.880 


EuLE  I. —  Write  the  parts  of  the  subtrahend  under  those 
of  the  same  denomination  in  the  minuend. 

Beginning  at  the  right,  subtract,  if  possiUe,  each  number 
in  the  subtrahend  from  that  above  it,  and  write  the  difference 
underneath. 

When  any  number  in  the  subtrahend  is  greater  than  that 
directly  above  it,  add  as  many  ones  to  the  upper  number  as 
maJce  1  of  the  next  higher  denomination. 

From  this  sum  subtract  the  lower  number,  and  add  1  to 
the  next  denomination  to  the  left,  before  subtracting. 

11. — Reduce  common  and  decimal  fractions  to  the  same 
denomination  ;  then  subtract  and  reduce  the  remainder  to  a 
denominate  number.    Or, 


127  cd.  106.880 

127.835  cd.  = 
cu.  ft. 

2d  solution. 
cd. 
129.716  cd.   =  129 
15.048  cd.  ft.  =   1 

cu.  ft. 

91.648 
112.768 

SUBTRACTION.  233 

Reduce  common  and  decimal  fractions  to  denominate  num- 
bers, and  then  subtract  according  to  Rule  I, 


ritOBLEMS. 

(1) 

( 

:2) 

(3) 

£   s.  d. 

qr. 

mi.-  fur. 

ch.  p. 

1. 

mi.  fur. 

ch. 

p.  1. 

^0   0  0 

0 

17  3 

5   1 

10 

23  4 

3 

2  15 

8  15  6 

3 

7  5 

7  3 

20 

14  6 

3 

1  18 

11   4  5 

1 

9  5 

7  1 

15 

8  6 

0 

0  22 

(4) 

(5) 

A.  R.  P. 

Sq.  ft. 

sq.  in. 

A. 

R.  P. 

yd. 

ft. 

in. 

19  1  25 

100 

101 

20 

2  20 

20 

5 

6 

4  3  30 

200 

130 

3 

3  30 

29 

6 

100 

14  1  34 

171(i) 

115 

16 

2  29 

20(1) 

7 

50 

(i) 

=  36 

(i)= 

=  2(i) 

16 

2  29 

21 

0 

)-36 

14  1  34 

172 

7 

86 

(6) 

(7) 

(8) 

cd.  cu.  ft 

).  cu.  in. 

cu.  yd. 

cu.  ft. 

cu.  in. 

T. 

cwt. 

lb. 

200  100 

1200 

20 

10 

1000 

500 

19 

27 

99  119 

1700 

9 

19 

1500 

37 

55 

48 

100  108 

1228 

10 

17 

1228 

461 

3 

79 

(9) 

(10) 

(11) 

lb.  oz.  pwt.  gr. 

lb. 

oz.  pwt.  gr. 

hr. 

min. 

sec. 

15  10 

10  12 

20 

23 

15 

14 

4  10 

15  20 

9 

9  : 

L9   9 

17 

41 

25 

10  11 

14  16 

10 

2 

0  15 

5 

33 

49 

234  DENOMINATE     NUMBERS. 


(13) 

(13) 

mi. 

fur.  rd. 

ft. 

in. 

mi. 

fur.  rd.  yd.   ft. 

in. 

125 

4  18 

11 

7 

18 

6   3  2    1 

9 

12 

6  23 

12 

9 

9 

6  28  3    1 

11 

112 

5  34 

14a) 

10 

8 

7  14  3^)   2 

10 

ay- 

=  6 

8 

a)=i 

7  14  4    1 

G 

112 

5  34 

lb 

4 

4 

14.  From  4ft)  7  i  take  2ft)  9  ?  5  3  13  15  gr. 

Ans.  1ft)  9  I  2  3  l3  5  gr. 

15.  From  2  ft)  4  ?  3  gr.  of  morphia,  a  druggist  sold  10  | 
4  3  l3  10  gr.    How  much  remained  ? 

Ans.  1ft)  5  §  3  3  l9  13gr. 

16.  From  2  T.  17  cwt.  15  lb.  of  sugar,  a  grocer  sold  19 
cwt.  75  lb.  9  oz.    How  much  remained  ? 

Ans.  1  T.  17  cwt.  39  lb.  7  oz. 

17.  From  a  barrel  containing  40  gal.  2  qt.  1  pt.  of  vinegar, 
were  sold  25  gaL  2  qt.  1  pt.  3  gi.    How  much  remained  ? 

Ans.  14  gal.  3  qt.  1  pt.  1  gi. 

18.  From  1  gal.  7  ^  of  laudanum,  a  druggist  sold  5  0 
13  f  3  3f3  40  n.    How  much  remained  ? 

Ans.  20  2fl  4f3  27^. 

19.  From  a  bin  containing  250  bu.  2  pk.  4  qt.  of  wheat, 
were  sold  190  bu.  3  pk.  5  qt.  1  pt.  How  much  remained 
unsold  ?  Ans.  59  bu.  2  pk.  6  qt.  1  pt. 

20.  If  the  moon  in  going  around  the  earth,  has  at  a  cer- 
tain time  gone  9  signs  22°  29'  31",  how  far  must  it  yet  travel 
to  make  a  complete  circuit?  Ans.  2  signs  7°  30'  29". 

21.  From  4  com.  yr.  and  100  da.  take  1  com.  yr.  289  da. 
11  hr.  19  min.  36  sec. 

Ans.  2  yr.  175  da.  12  hr.  40  min.  24  sec. 


SUBTRACTION.  235 

22.  Washington  is  38°  53'  39"  north  latitude,  and  Pitts- 
burgh is  40°  27'  36"  north  latitude.  What  is  their  difference 
of  latitude  ?  Ans.  1°  33'  57". 

23.  Washington  is  77°  2'  48"  west  longitude  from  Green- 
wich, and  Cincinnati  is  84°  29'  31"  west  from  Greenwich. 
What  is  their  difference  of  longitude  ?      Am.  7°  26'  43". 

24.  One  train  left  the  station  at  7  o'c.  45  min.  A.  M.,  and 
another  at  10  o'c.  25  min.  15  sec.  A.  M.  How  long  after 
the  first  did  the  second  start  ?     Ans.  2  hr.  40  min.  15  sec. 

25.  From  31  yd.  1  qr.  of  broadcloth,  were  cut  6  yd.  2  qr. 
How  much  remained  ?  Ans,  24  yd.  3  qr. 

26.  From  1  ream  10  quires  12  sheets  of  letter-paper,  were 
sold  15  quires  19  sheets.     How  much  remained? 

Ans.  14  quires  17  sheets. 

27.  From  a  great  gross  of  steel  pens,  were  sold  5  gross  5 
dozens  and  6.    How  many  remained  ? 

A71S.  6  gross  6  dozens  6. 

28.  From  £-^  take  ^s.  A71S.  4s.  4d.  2  far. 

29.  From  |  lb.  T.,  take  |  J  oz.     A71S.  9  oz.  18  pwt.  20  gr. 

30.  From  f  mi.  take  |  fur.    Ans.  1  fur.  7  rd.  1  ft.  10  in. 

31.  From  ^  A.  take  f  P. 

A71S.  2  R.  10  p.  29  yd.  6  ft.  133^  in. 

32.  From  |  wk.  take  ^  da.        A7is.  4  da.  5  hr.  48  min. 

33.  From  /j^  S.  take  ||°.  A7is.  5°  37'  6^"- 

34.  From  3^  T.  take  -^j  cwt. 

Ans.  3  cwt.  47  lb.  11  oz.  10^  dr. 

35.  From  7.35  T.  take  13.48  cwt. 

Ans.  6  T.  13  cwt.  52  lb. 

36.  From  10.37  f  1  take  3.416  f  3  . 

Ans.9fl  7f3  32.64ni. 

37.  From  4.75  lb  take  9.648  5  . 

Ans.  3ft).  11!  2  3  23  8.96  gr. 


236  DiEKOMINATE     NUMBEES. 

38.  From  3.87  oz.  T.  take  17.4  pwt.    Ans. 

39.  From  13.671  bu.  take  3.504  pk. 

Ans.  12  bu.  3  pk.  1  qt.  .88  pt. 

40.  From  .7  mi.  take  3  fur.  5  rd.  A7is.  2  fur.  19  rd. 

41.  From  17.316  A.  take  141.7  P. 

Ans.  16  A.  1  E.  28  P.  234.135  ft. 

To  find  the  time  hettveen  two  dates  in  the  same  era,  we  use 
the  denominations  years,  months,  and  days. 

For  the  fraction  of  a  year  this  method  is  not  exact ;  for  we  regard 
a  year  =  12  months,  and  a  month  =  30  days ;  hence  a  year  =  860 
days,  instead  of  365,  or  365|  days. 

Example. — What  is  the  difference  between  Mar.  3, 1876, 
andMay  11,  1867? 


Explanation. — We  write  the  number  of  the 
year,  month  and  day  of  each  date  in  order,  using 
the  later  date  for  the  minuend  and  the  earlier  for 
the  subtrahend. 


SOLUTION 

■. 

yr. 

mo. 

da. 

1876 

3 

3 

1867 

5 

11 

8       9       22 


EuLE. — Subtract  tlie  earlier  date  from  the  later,  reckon- 
ing  30  days  a  month,  and  12  months  a  year. 

Find  the  time  from 

42.  Nov.  15,  1812,  to  May  2,  1870. 

Ans.  57  yr.  5  mo.  17  da. 

43.  Oct.  9,  1784,  to  Aug.  1,  1875. 

Ans.  90  yr.  9  mo.  22  da. 

44.  Sept.  7,  1800,  to  July  4,  1846. 

45.  Dec.  21,  1790,  to  June  1,  1876. 

46.  Apr.  25,  1820,  to  Mar.  3,  1830. 

47.  Feb.  29,  1836,  to  Feb.  28,  1872. 

48.  Feb.  10,  1841,  to  Jan.  2,  1843. 


SUBTRACTION. 


237 


49.  Milton  was  born  Dec.  9,  1G08,  and  died  Nov.  8,  1675. 
How  old  was  he  ?  Ans,  66  yr.  10  mo.  29  da. 

50.  Money  borrowed  Oct.  22, 1861,  was  paid  Apr.  13, 1875. 
How  long  was  it  kept  ?  Ans.  13  yr.  5  mo.  21  da. 

In  calculating  Interest,  much  use  is  made  of  the  following 

TABLE. 

STwwing  the  number  of  days  from  any  day  of  one  month ,  to  the  same 
day  of  any  other  month  within  a  year  of  the  former. 


From  ant 

DAY  OF 

To  THE  SAME  DAT  OF  NEXT 

1^ 

u 

< 

i 

o3 

0 

5 

^ 

a 

^ 

o 
O 

1 

Q 

January 

365 

31 

59 

90 

120 

151 

181 

212 

243 

273 

304 

334 

February 

334 

365 

28 

59 

89 

120 

150 

181 

212 

242 

273 

303 

March 

306 

337 

365 

31 

61 

92 

12:3 

153 

184 

214 

245 

275 

April.-. 

275 

306 

334 

305 

30 

61 

91 

122 

153 

183 

214 

244 

May 

245 

376 

304 

335 

365 

31 

61 

92 

123 

153 

184 

214 

June  ...   ... 

214 

245 

273 

304 

334 

365 

30 

61 

92 

122 

153 

183 

July 

184 

215 

243 

274 

304 

335 

305 

31 

62 

92 

123 

153 

August 

153 

184 

212 

243 

273 

304 

334 

365 

81 

61 

92 

122 

September 

122 

153 

181 

212 

242 

273 

303 

334 

365 

30 

61 

91 

October 

92 

123 

151 

182 

212 

243 

273 

304 

335 

365 

31 

61 

November 

61 

92 

120 

161 

181 

212 

242 

273 

304 

334 

365 

30 

December 

31 

62 

90 

181 

151 

182 

212 

243 

274 

304 

335 

365 

1 

Example. — Find  the  number  of  days  from  Mar.  21  to 
Dec.  16. 

Solution. — We  find  Mar.  in  the  left-hand  column,  and  in  the  same 
line  in  the  column  headed  Dec,  we  find  275,  the  number  of  days  from 
any  day  in  Mar.  to  the  same  day  in  Dec.  Therefore,  from  Mar.  21  to 
Dec.  21  =  275  days.  But,  we  are  required  to  find  the  time  from  Mar. 
21  to  Dec.  16,  or  5  days  earlier  than  the  21st ;  we,  therefore,  subtract 
5  from  275,  and  find  270  days  as  the  true  time. 


-^ — «)©(^ 


mm\m:ikmiM\mMm\mn  m 


24A. 


:i'1>Yi^ittei\^7cferci^eX'{; 


Example. — What  cost  5  yards  broadcloth,  at  £1  10s.  6dc 
per  yard  ? 


SOLUTION. 

£      s.     d. 
1     10     6 

5 

7    12     6 


Explanation. — Since  1  yd.  costs  £1  10s.  6d., 
5  yd.  will  cost  5  times  as  much. 

5  times  6d.  =  80d.  =  2s.  6d.  Write  6  in  the 
pence  place  of  the  product,  and  reserve  the  2s.  to 
be  added  to  the  product  of  the  shillings. 

5  times  10s.  =  50s.,  to  which  add  the  2s.,  making 
52s.  =  £2  12s.     Write  12  in  the  shillings'  place 
of  the  product,  reserving  the  £2  to  add  to  the  product  of  pounds. 

5  times  £1  =  £5,  to  which  add  the  £2  reserved,  making  £7.  The 
entire  product,  therefore,  is  £7  12s.  6d. 

EuLE. — Multiply  the  number  represe7iting  the  denomina- 
tion at  the  right  by  the  multiplier.  Divide  the  product  by 
the  number  of  this  denomination  required  to  make  1  of  the 
next  higher.  Write  the  remainder,  if  any,  ifi  the  product 
under  the  number  multiplied,  and  add  the  quotient  to  the 
product  of  the  next  higher  de7iomination. 

Proceed  in  this  manner  till  all  the  denominations  have 
been  multiplied. 

The  result  will  be  the  answer  sought. 


MULTIPLICATION.  239 


PJtOBZTlMS. 

(1-) 

(2.) 

mi.  fur.  rd.     ft. 

in.                           mi.fur.  cli.   p. 

1. 

5     7     13     10 

8                          2     6     7     3 

15 

5 

4 

29    4    28      3(i)     4  11     3     1     2     10 

a)=6 

29    4    28      3         10 


(3.) 

(4.) 

A.    R.     P.       ft. 

in. 

A.    R.    P.     yd.    ft.    in. 

8     2     16     138 

100 

7    3     27     15     6     50 

2 

3 

17    0    33 

4(J)       56 
(|-)=108 

17     0     33 

5            20 

23     3       2     16}  1       6 
i=6  108 

23     3       2     16     7  114 


(5.)  (6.)  (7.) 


ou.yd.cu.ft.  cu.in. 
6       20      435 

cd.cd.ft. 
5     7 

cu.ft. 
13 

cd.  cu.  ft.  cu.  in. 
2      52      360 

4 

59      6 

10 
2 

5 

23         0         12 

12       5        72 

(8.) 
lb.  oz.pwt.gr. 
6     7     8     9 

C.   0. 

4 

(9.) 
tl    il 
13     6 

lb. 

1 

(10.) 

!  3  3  gr. 
10     7     1     10 

6 

9 

7 

39     8  10     6  5    3     11     6  13      4    4    1     10 

11.  What  is  the  weight  of  8  bales  of  hay,  each  weighing 
5  cwt.  6  lb.  ?  Ans,  2  T.  48  lb. 


240  DETS^OMINATE     N^  UMBERS. 

12.  What  is  the  weight  of  10  packages  of  crockery,  aver- 
aging in  English  Avoirdupois  weight,  7  cwt.  26  lb.  7  oz. 
9  dr.  ?  Ans.  3  T.  12  cwt.  40  lb.  11  oz.  10  dr. 

13.  Find  the  contents  of  9  bbl.,  averaging  40  gal.  1  qt 

1  pt.  1  gi.  Ans.  363  gal.  2  qt.  1  pt.  1  gi. 

14.  How  much  castor-oil  in  50  bottles  averaging  3  f  ^ 

2  f  3  ?  Ans.  1  Cong.  2  0.  2  f  1  4  f  3  . 

15.  How  much  wheat  in  19  sacks,  averaging  1  bu.  3  pk. 
7  qt.  1  pt.?  Ans.  37  bu.  2  pk.  6  qt.  1  pt. 

16.  Multiply  75°  45'  25"  by  7.     Ans.  1  0.  170°  17'  55". 

17.  Calling  a  solar  year  365  da.  5  hr.  48  min.  50  sec,  how 
much  time  is  there  in  18  solar  years? 

Ans.  6574  da.  8  hr.  39  min. 

18.  How  much  molasses  in  25  casks,  if  each  cask  contains 
61  gal.  1  qt.  1  pt.  ?  Ans,  1534  gal.  1  qt.  1  pt. 

19.  If  one  silver  spoon  weighs  3  oz.  15  pwt.  13  gr.,  what 
is  the  weight  of  two  dozen  ?         Ans.  7  lb.  6  oz.  13  pwt. 

20.  If  a  man  travels  at  the  rate  of  22  m.  7  fur.  32  rd. 
4  yd.  per  day,  how  far  can  he  travel  in  56  days  ? 

Ans.  1286  m.  5  fur.  32  rd.  4  yd. 

21.  How  many  bushels  of  wheat  will  12  acres  produce, 
if  one  acre  produces  30  bu.  3  pk.  7  qt.  1  pt.  ? 

Ans.  371  bu.  3  pk.  2  qt. 

22.  How  much  cloth  will  it  take  for  16  suits  of  clothes, 
if  it  takes  8  yd.  If  qr.  for  one  suit  ?  Ans.  135  yd. 

23.  If  one  load  of  wood  measures  117  cu  ft.  110  cu.  in., 
how  much  will  40  loads  of  the  same  size  measure  ? 

Alls.  36  cd.  74  cu.  ft.  944  cu.  in. 

24.  If  a  quarter  of  beef  weighs  216  lb.  7  oz.,  how  mu(*h 
will  4  quarters  weigh  ?  Ans.  865  lb.  12  oz. 

25.  If  the  daily  motion  of  the  moon  is  13°  10'  35",  how 
much  is  it  in  15  days  ?  Ans.  197°  38'  45". 


^ 


^iv 


I>;l^IS;iO;]^ 


245. 


t^\¥i?itteri  ^x:Grci;^eX 


Example  1.— If  5  yd.  of  broadcloth  cost  £7  12s.  6d.,  what 
^osts  1  yard  ? 


1st  solution. 

£     s.    d. 
I  of  £7    =  1     8     0 
I  of  12s.  =        2     4f 
i  of  6d.  = H 

1  10     6 


1st  Explanation. — Since  5  yd. 
cost  £7  12s.  6d.,  1  yard  will  cost  | 
as  much. 

^  of  £7    =  £lf  =  £1  8s. 

^  of  12s.  =  2|s.  =  2s.  4fd. 

iof6d.    =lid. 
£1  8s.+2s.  4fd.  +  Ud.=£l  10s.  6d. 


2d  solution. 
£      s.      d. 
7    12     6  =  £7.625 

5  )  £7.625 

£1.525  =  £1  10s.  6d. 


2d  Explanation. — £7  12s.  6d.= 
£7.625. 

Dividing  by  5,  we  have  £1.525  = 
£1  10s.  6d. 


3d  solution. 
£      s.      d. 
7    12     6  =  1830d. 

5 )  1830d. 


366d.  =  £1  10s.  6d. 
16 


3d  Explanation.— £7  12s.  6d.= 
1830d. 

Dividing  by  5,  we  obtain  366d.  = 
£1  10s.  6d. 


S41 


242 


DEKOMIHATE     I^^UMBEES 


Example  2.— At  £1  10s.  6d.  per  yd.,  how  many  yards  of 
broadcloth  can  be  bought  for  £7  12s.  6d.  ? 


1st  solution. 
£      8.     d.    £      s.     d. 
1     10     6)7     12     6(5 
7     12     6 


1st  Explanation.  —  By  inspec- 
tion we  see  that  £7  12s.  6d.  contains 
£1  10s.  6d.  exactly  5  times. 


3d  solution. 
£      s.      d.  £ 

7     12     6  =  7.625 
1     10     6  =  1.525 

£7.625  -r-  £1.525  =  5 


2d  Explanation.  —  We  reduce 
both  dividend  and  divisor  to  £  and 
the  decimal  of  a  £,  and  divide  as  in 
division  of  decimals. 


8d  solution. 
£      s.     d.         d. 
7     12     6  =  1830 
1     10     6  =    366 

1830d.  -T-  366d.  =  5 


3d  Explanation.  •  —  We  reduce 
both  numbers  tod.,  and  then  divide 
as  in  whole  numbers. 

Hence, 


To  find  any  required  part  of  a  simple  or  compound  denominate 
number,  we  have 

KuLE  I. — Find  the  required  part  of  each  simple  denom- 
inate number  composing  the  dividend,  reducing  these  parts, 
if  necessary,  to  compound  denominate  numbers.  Then  add 
as  in  addition  of  denominate  numbers.    Or, 

Reduce  the  dividend  to  any  convenient  denominate  number, 
and  proceed  as  in  division  of  integers,  or  decimals.  Then 
reduce  the  result,  if  desirable,  to  a  denominate  number. 

If  we  cannot  by  inspection  divide  one  compound  denominate 
number  by  another  we  apply 

KuLE  II. — Reduce  dividend  and  divisor  to  the  same 
denomination,  and  divide  as  in  integers,  decimals,  or  frac- 
tions. 


DiYisioiq^.  243 


PJt  O  B  LEMS 


(1)  (^)  (3) 

mi.   fur.  rd.                     mi.     fur.  rd.                       A.      R.  P. 

5)29     7     8             4)23     52     6              9 )  35     1  12 

3     3  28 


5 

7  33.6 

rd. 
6)75 

(4) 

yd. 

4 

ft.  in. 

2     6 

c 
9 

12 

3 

1     8 

11 

lb. 

)35 

(7) 
oz.  pwt. 
2     3 

10 

3 

2     7 

14 

5 

19 

li 

1 

(5) 

cu.  yd 

.ft. 

in. 

9)30 

4 

9 

3 

9 

769 

(6) 

cd.     ft.  in. 

10 )  93     12  632 

9     39  1100 


(8) 

lb.     1      3  3  gr. 

3)11     1     6  1  17 

3     8     4  2  12J 


(9)  (10) 

T.   cwt.  qr.    lb,     oz.      dr.  gal.   qt.   pt.   gi. 

5)53     4     1     19     14     12  12 )  47     1     1     2 


10  12     3     13     15     12  3     3     1     2i 

11.  If  9  bales  of  hay  weigh  2  T.  7  cwt.  25  lb.,  what  is  the 
average  weight  ?  Ans.  5  cwt.  25  lb. 

12.  If  10  packages  of  crockery  weigh  in  long  ton  weight, 
3  T.  14  cwt.  3  qr.  12  lb.  11  oz.  10  dr.,  what  is  their  average 
weight?  Ans.  7  cwt.  1  qr.  26  lb.  7  oz.  9  dr. 

13.  If  9  barrels  hold  363  gal.  3  qt.  1  pt.  2  gi.,  what  is 
their  average  capacity  ?  Ans.  40  gal.  1  qt.  1  pt.  2  gi. 

14.  If  10  sacks  contain  39  bu.  2  pk.  6  qt.  of  wheat,  what 
is  their  average  capacity  ?  Ans.  3  bu.  3  pk.  7  qt. 

15.  How  many  town  lots  containing  28  P.  30  yd.  each, 
can  be  made  of  3  A.  2  R  19  P.  25.25  yd.  ?  Ans.  20. 


244 


DENOMINATE    NUMBERS, 


OUTLINE   OF  APPLICATIONS   OF  SQUARE,   CUBIC, 
AND   TIME   MEASURES. 


O 

M 

H 

M 

-^ 


SQUARE. 


CUBIC. 


247.  Common, 

248.  Government  Lands. 

249.  Lumber. 
C  250.  Plasterers'  Work. 

251.  Carpenters'  Work. 

252.  Painters'  Work. 
I  253.  Paviors'  Work. 


Artificers' 
Work, 


254.  Excavations  and  Embank- 
ments. 


Masonry, 


255.  Stone  Work. 


256.  Brkk  Work. 
^  257.  Capacities, 


258.  TIME. 


APPLICATIOKS.  245 


APPLICATIONS    OF    SQUARE,    CUBIC,    AND    TIME 
MEASURES. 

346.  We  propose  in  this  chapter  to  give  some  such 
practical  applications  of  Square,  Cubic,  and  Time  Measures, 
as  the  pupil  is  now  prepared  to  understand ;  omitting  for 
the  present,  those  that  involve  square  root,  cule  root,  &c. 
These  latter  will  be  treated  of  in  another  place. 

347.  SQUARE   31EASUBE. 

Example  1. — How  many  sq.  yd.  in  the  floor  of  a  rect- 
angular room  5  yd.  long  and  4  yd.  wide  ? 

Solution. — Since  a  floor  5  yd.  long  and  1  yd.  wide  contains  5  sq.  yd., 
a  floor  5  yd.  long  and  4  yd.  wide  contains  4  times  5  sq.  yd.,  or  20  sq.  yd. 
Hence  to  find  the  area  of  a  rectangular  surface  (204),  we  have  this 

Rule. — Multiply  the  length  by  the  width,  expressed  in  the 
same  denomination. 

F  HOB  LEMS. 

1.  How  many  sq.  in.  in  a  rectangular  pane  of  •  glass 
16  X  18  in.  ?     How  many  sq.  ft.  ?     What  part  of  a  sq.  yd.  ? 

2.  How  many  sq.  ft.  in  a  box  of  glass  containing  60  panes, 
each  10  x  12  in.  ?  Ans.  50  sq.  ft. 

3.  How  many  perches  in  a  village  lot  150  ft.  long  and 
50  ft.  wide?  27.55+  P. 

4.  How  many  acres  in  a  square  farm,  eacli  of  whose  sides 
is  40  rd.  Ans.  10  A. 

5.  How  many  acres  in  a  square  piece  of  land  measuring 
G  miles  on  each  side?  Ans.  23040  A. 


246  DENOMINATE     NUMBERS. 

Example  2. — If  a  rectangular  pane  of  glass  18  in.  long, 
has  a  surface  of  288  sq.  in.,  how  wide  is  it  ? 

SOLUTION.  Explanation. — Since  a  pane  of  glass  18  in.  long  and 

18  )  288       1  in-  wide  contains  18  sq.  in.,  to  contain  288  sq.  in.  it  will 
Tn       liave  to  be  as  many  inches  wide  as  18  sq.  in.  are  con- 
tained times  in  288  sq.  in.,  or  16  inches  wida. 
Hence,  to  find  one  dimension  of  a  rectangle  (207),  when  the  other 
dimension  is  given,  we  have  this 

EuLE. — Divide  the  area  hy  the  giveti  dimension  ;  the  quo^ 
tient  is  the  other  dime7ision, 

6.  A  town  lot  contains  2400  sq.  ft.    It  has  20  ft.  front. 
What  is  its  deptfi  ? 

7.  A  box  of  glass  contains  50  sq.  ft.  How  many  panes 
12  X  24  in.  ?  Ans.  25  panes. 

8.  How  many  rods  in  length  is  a  rectangular  farm  con- 
taining 150  acres,  if  one  end  is  120  rd.  in  width  ? 

9.  How  many  yards  of  carpet  -|  yd.  wide,  will  cover  a 
floor  15  feet  square  ?  Ans.  40  yd. 

10.  A  gentleman  purchased  Brussels  carpet  |  yd.  wide,  at 
$1.75  per  linear  yard,  to  cover  a  room  22  feet  long.  The 
entire  cost  of  carpet  was  $106.08f.  What  was  the  width  of 
the  room  ?  Ans.  15.5  ft. 

11.  *How  many  yards  of  oil-cloth  will  be  required  to  cover 
a  hall  25  feet  long  and  9J  feet  wide;  and  two  recesses  each 
6  X  7  ft.  ? 

12.  How  much  will  it  cost  for  blinds  for  10  windows,  each 
7x3  ft.,  at  62i^  per  yd.  ?  Ans.  $14.58  + . 

248.  Government  Zand  Measure, 

1.  If  in  a  Section  there  are  640  acres  of  land,  how  many 
acres  are  there  in  the  E.  |  of  S.  |^  of  Section  10  ?  (See  Dia- 
gram, page  191.) 


APPLICATION'S.  247 

2.  How  many  in  the  S.  E.  J  of  S.  E.  J  of  Sec.  10  ? 

3.  How  many  in  the  E.  J  of  E.  ^  of  S.  W.  J  of  Sec.  21  ? 

4.  How  many  in  the  W.  ^  of  E.  ^  of  S.  W.  i  of  Sec.  36  ? 

5.  A  farmer  purchased  the  N.  W.  J  of  Sec.  1,  Township  6, 
N.,  Range  G  E.  of  the  oth  Principal  Meridian ;  and  sold  35 
acres  of  it  to  his  brother.     How  many  acres  had  he  left  ? 

Ans,  125  A. 

6.  If  he  paid  the  usual  Government  pre-emption  price 
($1.25  per  acre),  and  sold  to  his  brother  at  the  rate  of  13.25 
per  acre,  what  did  the  part  retained  stand  him  ? 

Ans.  $86.25. 

7.  If  he  afterwards  exchanges  his  property  for  S.  ^  of  the 
S.  W.  J  of  Sec.  30,  in  the  same  Township,  how  much  per 
acre  does  his  new  land  cost  him  ?  Ans.  $1,078^. 

Show  on  the  diagram  the  position  of  each  purchase. 

8.  A  speculator  bought  the  K  i  of  the  E.  |-  of  the  S.  W.  i 
of  Sec.  8 ;  he  afterwards  purchased  the  N.  ^  of  the  E.  ^  of 
the  N.  E.  i,  the  S.  i  of  the  W.  J  of  the  N.  E.  J,  and  the 
W.  J  of  the  S.  i  of  the  S.  W.  i  of  the  same  section.  How 
much  land  did  he  purchase  ?  Ans.  160  acres. 

Make  a  diagram  showing  the  shape  of  his  land. 

9.  He  afterwards  made  an  even  exchange  of  two  of  the 
pieces  he  had  purchased,  so  that  his  property  was  in  the 
form  of  a  single  square.  Show  in  how  many  ways  this  may 
be  done. 

349.  Lumber  Measure, 

Lumber  is  reckoned  by  ioard  measure;  that  is,  it  is 
regarded  as  cut  into  boards  one  inch  thick.  The  unit  of 
board  measure  is  a  square  foot,  one  inch  thick.  If  a  piece 
of  lumber  1  inch  thick,  contains  a  certain  number  of  feet,  a 
piece  with  the  same  dimensions  ttvo  inches  thick,  contains 


248  DEKOMIKATE     NUMBERS. 

twice  as  many  feet ;  three  inches  thick,  three  times  as  many 
feet,  &c. 

The  average  width  of  a  board  is  half  the  sum  of  the  width  of  the 
ends. 

Example  1. — How  many  sq.  ft.  in  a  board  18  ft.  long, 
18  in.  wide,  and  1  in.  thick  ? 

Explanation. — We  reduce  the  width 

SOLUTION.  18  in.  to  1.5  ft.  by  dividing  by  13;  then 

18  in.  =:  1.5  linear  ft.     multiply  1.5  by  J  8,  and  thus  obtain  27,  to 

-tn        a       a       which  we  give  the  name  sq.  ft.     As  the 

board  is  1  in.  thick,  the  surface  is  the  pro- 

27  sq.  ft.  duct  of  the  length  and  breadth,  the  true 

result  being  27  sq.  ft. 


Example  2. — How  many  feet  in  a  plank  18  ft.  long, 
18  in.  wide,  and  1^  in.  thick  ? 

SOLUTION.  Explanation. — We  proceed,  as 

.14  X  18  X  IJ  =  40|-  sq.  ft.     in  the  last  example,  to  find  the  area. 

Since  the  plank  is  1^  times  as  thick 
as  the  board,  it  must  contain  1\  times  as  many  feet     1^  times  27=40|. 

Rule. — Multiply  together  the  length  and  breadth  (or  aver- 
age Ireadth)  in  feet ;  and  this  product  ty  the  thickness  in 
inches. 

PIIOBI.EMS, 

1.  How  many  feet  in  an  inch  board  10  ft.  long  and  15  in. 
broad  ?  Ans.  l^  ft. 

2.  How  many  feet  in  a  three-inch  plank  10  ft.  long  and 
15  in.  broad  ?  Ans.  37^  ft. 

3.  In  a  log  12  ft.  long  and  15  in.  square?    Ayis.  225  ft. 

4.  In  a  joist  20  ft.  long,  10  in.  broad,  and  3  in.  thick? 

Ans.  50  ft. 

5.  In  a  scantling  16  ft.  long  and  4  in.  square  ? 

Ans.  21ift. 


APPLICATIONS.  ^  249 

6.  In  a  beam  30  ft.  long,  8  in.  broad,  and  6  in.  thick  ? 

Ans.  120  ft. 

7.  What  is  the  value,  at  2J^  per  foot,  of  a  log  25  ft.  long, 
18  in.  broad  at  one  end,  14  at  the  other,  and  uniformly  14 
in.  thick?  Ans.  $11.67. 

8.  What  is  the  value  of  a  plank  30  ft.  long,  2  ft.  wide,  H 
in.  thick,  at  3^^  per  foot,  board  measure  ?         Ans.  $3.15. 

9.  What  cost  37  joists,  each  20  ft.  long,  9  in.  broad,  and 
2^  in.  thick,  at  $18  per  1000  feet  ?  Ans.  $24.98. 

One  thousand  shingles  4  in.  wide  are  estimated  to  cover  one  square 
of  100  sq.  ft. 

10.  How  many  shingles  will  be  required  to  cover  a  roof, 
the  one  side  of  which  is  40  by  20  ft.,  and  the  other  40  by 
18  ft.  ?  Ans.  15i  M. 

ARTIFICERS'    WORK. 

250.    Plasterers^  Work, 

Plastering  is  done  by  the  square  yard. 

Rules  for  Measuring.  I. — Girt  {measure)  the  walls 
and  multiply  hy  the  height  from  floor  to  ceiling. 

Note  1. — Deduct  the  half  of  all  openings  over  two  feet  wide. 

2.  Measure  all  circular  walls  and  ceilings  twice. 

3.  Corners  of  chimneys  and  other  external  angles  are  charged  extra 
if  made  with  "  gauged  "  mortar,  otherwise  they  are  included  in  the 
common  measurement. 

4.  Stucco  work  is  done  by  the  square  foot,  when  more  than  13  in. 
wide. 

II. — Girt  the  cornice  and  multiply  hy  the  length, 

5.  One  foot  in  length  of  the  cornice  is  added  for  each  mitre  oi' 
"return." 

Example  1. — At  30^  per  yd.,  what  will  it  cost  to  plaster 
a  room  16^  ft.  long,  13^  ft.  broad,  and  10  ft.  high,  having  a 


250  DEKOMIN^ATE     NUMBERS. 

door  7J  ft.  by  3  ft.,  two  windows  each  6  ft.  by  3,  and  a 
mantel  5  ft.  by  4  ? 

SOLUTION. 

16|  +  16|  +  13|  +  13|  =  60,  the  measurement  around  the  room. 
60     X  10    =  600,  the  number  of  sq.  ft  in  the  walls. 
16^  X  131  =  222|,  "        "         "        "      "     •'    ceiling. 
822|,  "        "         *•        "      «    "    room. 

7^  X  3  =  22|,  the  number  of  sq.  ft.  in  the  door. 
2  X  6    X  3  =  36  ,    "        "         "       .<     «     «    two  windows. 
5x4  =  20_,    "         "  "        "      "    "    mantel. 

78^,    *'        "         "        "      "     "    openings. 

78^  H-  2  =  391,  the  number  of  sq.  ft.  to  be  deducted  (Note  1). 
822f  -  39^  =  783i  sq.  ft.  =  87tV  sq.  yd. 
87yV  sq.  yd,  @  30;^  =  $26.12,  Ans. 

Example  2. — What  cost  cornice  15  in.  wide,  about  the 
same  room,  supposing  the  only  mitres  to  be  those  at  the 
corners,  at  10^  per  foot  ? 

SOLUTION. 

Girth  of  room,  60  ft.  ;  4  mitres,  4  ft. ;  60  ft.  +  4  ft.  =  64  ft. ;  15  in. 
=  1\  ft. ;  64  X  1^  =  80  sq.  ft. 
80  sq.  ft.  @,  lOj^  =  $8.00,  Ans. 

PROBZEM8. 

1.  What  cost  the  plastering  of  a  circular  wall  63  ft.  in 
length  and  13  ft.  higb,  at  25)*  per  yd.  ?     (See  Note  2.) 

Ans.  $45.50. 

2.  How  many  yd.  of  plastering  will  cover  a  ceiling  20  x  30 
ft. ;  and  what  will  it  cost  at  15f  per  sq.  yd.  ? 

Ans.  GQ^  yd. ;  $10. 

3.  What  cost  the  plastering  of  the  walls  and  ceiling  of  a 
room  12  ft.  11  in.  square  and  9  ft.  6  in.  high,  at  20^  per  sq. 
yd.?  Ans,  $14.62  +. 


APPLICATIONS.  251 

4.  What  cost  the  cornice  of  a  room  24  x  16  ft.;  the 
room  having  two  chimneys,  breasts  1  ft.  thick,  and  forming 
4  extra  mitres  and  four  "returns; "  the  width  of  the  cornice 
as  measured  by  the  line  following  all  the  indentations  being 
16  inches ;  at  12J^^  per  foot  ?  Ans.  $16. 

5.  What  cost  the  plastering  of  a  house  of  8  rooms,  4  of 
which  are  18  x  15  ft.  and  8^  ft.  high ;  4  are  14  x  15  ft.  and 
9^  in  height,  and  a  hall  30  x  8  ft.  and  9^  ft.,  allowance 
being  made  for  16  doors,  7^  x  3,  and  18  windows  6^  x  3, 
at  15;^  per  sq.  yd.  ?  ^ws.  $116.24. 

251.  Carpenters^  and  Joiners^  Work. 

This  work  is  done  by  the  lineal  or  square  foot,  or  by  the 
square  of  100  sq.  ft. 

Framing  the  larger  timbers  used  in  building,  such  as  sills,  posts, 
and  vertical  timbers,  principal  rafters,  &c.,  framing  hips  and  valleys, 
chamfering  posts  and  girders,  raising  plates,  bond  timbers,  bridging 
joists,  making  mouldings,  architraves,  &c.,  are  counted  by  the  lineal 
foot. 

Planing  posts  and  girders,  making  window-shutters,  doors,  wain- 
scoting, pews,  shelving  (sometimes  lineal),  ceilings,  lattice-worR, 
cornice,  &c.,  are  reckoned  by  the  sqitnre  foot. 

Framing  joists,  small  rafters,  studding,  putting  on  weather-boarding, 
shingling,  sheathing,  laying  floors,  making  partitions,  board  fences, 
&c. ,  are  charged  for  by  the  square. 

Note  1.— In  measuring  weather-boarding,  deduct  all  openings  over 
4  ft.  wide. 
2.  Window-sashes  are  estimated  by  the  light. 

PBOBZEMS. 

1.  At  $5  a  square,  what  will  it  cost  to  shingle  a  roof 
45  ft.  long,  each  of  its  two  slopes  being  20  ft.  broad  ? 

Ans.  $90. 


252  DENOMINATE     NUMBEES. 

2.  At  50^  a  sq.  yd.,  what  costs  a  wainscot  2 J  ft.  high  and 
54  feet  in  length  ?  ^7i5.  $7.50. 

3.  What  will  a  floor  33  ft.  long  and  17  ft.  broad  cost,  at 
13.50  a  square  ?  Ans.  $19M. 

4.  What  cost  the  framing  of  4  sills,  6x8  in.  and  20  ft. 
long,  @  Sf  per  lineal  foot  ?  Ans.  $6.40. 

5.  What  cost  the  making  of  8  doors,  8x3  ft.,  1 J  in.  thick, 
6  panels,  at  16^  per  sq.  ft.?  Ans,  8 

6.  What  cost  the  making  of  20  shutters  each  18  in.  by 
6  ft.,  at  15^  per  superficial  foot  ?  Ans.  127. 

7.  What  will  be  the  cost  of  a  gravel  roof  20  x  30  ft.,  at 
$5.50  per  square  ?  Ans.  $ 

8.  What  costs  the  same  roof  of  shingles  @  $9.70.  Of  slate 
@  $11.     Of  tin  at  $16  per  square  ?         A7is.  to  last,  $96. 

9.  A  square  of  slate  roofing  weighs  600  lb.  What  will  be 
the  weight  of  slate  on  the  roof  in  the  last  problem  ? 

Ans.  3600  lbs. 

352.  Painters^  Work, 

I^riming  and  glaring  are  done  by  the  light. 
•  Painting  is  done  either  by  the  lineal  foot,  or  square  yard, 
and  is  estimated  as  any  other  surface  measure,  hy  the  jjro- 
duct  of  the  length  and  Ireadth,  The  principal  thing  is  to 
know  how  to  obtain  these  two  dimensions;  and  for  this 
purpose  we  give  the  following 

Directions  for  Measuring, 

In  weather-boarding,  measure  not  only  the  width  of  the  board,  but 
also  the  width  of  the  exposed  edge.  For  beaded  partition  boards, 
add  to  the  width  of  the  boards  |  inch  for  each  bead.  These  measure- 
ments multiplied  by  the  length  of  the  boards  give  the  area  painted. 

For  paling  fences,  measure  the  height,  and  to  this  add  the  width  of 
each  rail,  to  obtain  one  dimension,  breadth.    The  other  dimension, 


APPLICATIONS.  253 

^,  is  found  by  adding  to  the  length  of  the  fence,  twice  the  width 
of  each  post,  if  rectangular,  or  one-half  the  girth,  if  cylindrical.  [The 
measure  of  width  of  rails  and  posts  is  made  at  right  angles  with  the 
fence.] 

To  measure  panel  doors,  shutters,  &c.,  &c.,  find  the  length  and  breadth 
by  pressing  the  measuring  line  into  all  the  quirks  and  mouldings. 

For  all  weather- boarding  and  wall  painting,  take  out  half  the 
openings. 

For  shingle  or  board  roofs,  measure  the  butts  or  edges  ;  trellis  or 
lattice  work  to  be  measured  double,  and  Venetian  shutters  add 
one-half. 

To  measure  pilasters,  commence  at  the  wall  and  measure  round  to 
the  bead  at  the  end  of  the  jamb  casing,  pressing  the  line  into  all  the 
quirks.     This  gives  one  dimension  ;  the  other  is,  of  course,  the  length. 

Plain  rosettes,  in  pilasters,  add  1  foot  to  the  length  ;  carved 
rosettes,  in  pilasters,  add  3  feet  to  the  length. 

Window  jambs,  cornice,  &c.,  &c.,  are  measured  substantially  in  the 
same  way,  the  object  being  to  obtain  the  surface  painted,  whether  it 
be  plain  or  ornamental. 

mOBTj  E  MS  , 

1.  What  will   the  glazing  cost  in  a  house   containing 

17  windows  of  6  lights  each,  at  25^  a  light  ? 

2.  What  will  it  cost  to  paint  both  sides  of  a  paling  fence. 

18  rd.  long,  and  5  feet  high,  the  posts  4  x  4  in.  and  2  to  the 
rod,  and  rails  2  x  4in.  at  21^  per  sq.  yd.?   Ans.  I85.06  +  . 

3.  What  cost  the  painting  of  4  ft.  high  wainscoting  of 
6  in.  wide  beaded  boards,  of  a  room  15  x  17  ft.,  at  20^  per 
sq.  yd.?  Ans.  $6.16  +  . 

253.  Paviors^   Work. 

PliOBLEMS, 

1.  What  will  be  the  cost  of  flagging  a  walk  100  ft.  long 
by  9  ft.  wide,  at  21f  per  sq.  ft.  ?  Ans.  $243. 

2.  At  $6  per  square,  what  will  it  cost  to  pave  a  yard 
150  ft.  long  and  100  ft.  broad  ?  Ans.  $900. 


254  DEN^OMIJS^ATE    NUMBERS. 

3.  What  will  it  cost  to  pave  a  street  30  ft.  wide  and 
1560  ft.  long,  with  Nicholson  pavement,  @  $.30  per  sq.  ft.? 

^  Ans.  $ 

4.  What  cost  the  paving  of  an  avenue  100  ft.  wide  and 
3150  long  with  the  Miller  pavement  at  44^  per  sq.  yd.  ? 

Ans.  $15400. 

CUBIC  MUASUBE. 
254.  JExcavations  and  Enibankments, 

Excavations  and  Enihanknients  of  all  kinds  are 
reckoned  by  the  cubic  yard. 

Note. — Trenches  for  foimJations  are  counted  double. 

Example. — How  many  yards  of  excavation  in  a  cellar 

40  ft.  by  25  ft.  and  ^  ft.  deep  ? 

Solution. — Since  the  solid  contents  of  a  rectangular  body  are 
found  by  multiplying  together  its  three  dimensions  (length,  breadth, 
and  thickness),  we  multiply  40  by  25,  and  this  product  by  4J^,  obtain- 
ing as  a  result  4500  cu.  ft.  Dividing  this  by  27,  the  number  of  cu.  ft. 
in  1  cu.  yd.,  we  have  166|,  the  number  of  cu.  yd.  of  excavation. 

rjt  OB1.EMS, 

1.  What  cost  the  excavation  of  a  trench  for  a  foundation 
for  a  stone  fence  500  ft.  long,  2  ft.  wide,  and  1^  ft.  deep,  at 
27/  per  cu.  yd.  ?     (See  Note.)  Ans.  $30. 

2.  How  many  loads  of  1  cu.  yd.  each  will  be  required  to 
fill  a  street  150  ft.  long,  50  ft.  wide,  and  2^  ft.  deep;  and 
how  much  will  the  whole  cost,  at  18/  per  cu.  yd.  ? 

Ans.  694f  yds.  ;  $125. 

3.  A  railroad  company  excavated  a  tunnel  1500  feet  long, 
with  a  cross-section  of  450  square  feet.  How  many  yards 
of  earth  were  removed  ?  Ans.  25000  cu.  yd. 


APPLICATIONS.  255 

4.  What  costs  the  excavation  for  a  cellar  5  ft.  deep,  for  a 
dwelling  house,  the  main  building  being  40  x  30  ft.,  and  the 
L  18  X  14  ft.,  at  50)^  per  cu.  yd.  ?  Ans.  $134.44. 

5.  What  costs  the  excavation  for  a  cellar  under  the  main 
building  of  same  dwelling-house,  and  an  excavation  H  ft. 
deep  and  1^^  ft.  wide  for  the  walls  of  the  L  (one  of  the  short 
sides  being  attached  to  the  main  building),  at  50^  per 
cu.  yd.  ?  Ans,  $115.03. 

MASOJVBT. 
255.  Stone  Work. 
Stone  Worh  is  done  by  the  perch  of  24|  cu.  ft. 

DIRECTIONS   FOR   MEASURING. 

Rough   Stone. 

1.  Measure  the  outside  girt  of  a  wall.  This  measures  the  comers 
twice ;  but  as  a  corner  is  more  diflBcult  to  build,  this  is  a  method  of 
allowing  for  the  extra  work. 

2.  Add  9  in.  in  chimney  breasts,  pillars,  &c.,  for  each  dressed  and 
plambed  corner. 

3.  Count  openings  less  than  3  ft.  wide  as  built  up. 

Cut   Stone. 

1.  In  cut  stone  the  measuring  line  follows  the  chisel,  and  the  work 
is  measured  on  the  surface  by  the  square  foot. 

2.  Belt  courses  less  than  1  ft.  wide  are  measured  by  the  linear  foot. 

3.  Moulded  and  chamfered  work  and  the  whole  face  of  the  rise  in 
moulded  steps,  is  measured  twice. 

4.  Circular  moulded  hoods,  pediments,  and  lintels  are  measured 
twice. 

Example. — What  is  the  cost  of  a  foundation  36  x  20  ft, 
4  ft.  deep,  and  H  ft.  thick,  built  of  rubble  work  with  11  in. 
belt  course  of  cut  stone,  at  $3.63  per  perch  for  the  rubble, 
and  25^  per  foot  lineal,  for  the  belt  course? 


256  DENOMINATE     NUMBERS. 

SOLUTION.  EXPLANATIOX.— Mul- 

36  4-  20  4-  36  +  20  =  112  ft.,  girth.      tiplying  the  girth  by  the 
112  X  4  X  li  =r  672  cu.  ft.  ^®P*^'  ^°<l  *1^^*  product 

ci^-o    .    siA^        ow  's  1,  by  the  width  of  the  wall, 

672 -24f^27A  perch.         /e  find  673  cu.  ft. 

$3.63  X  27^  =  $98.56.  Dividing  this  by  24|, 

$.25  X  112    =  128  *°  ''^^^'^®  *^  perches,  we 

have  27/3  perches,which, 

$126.56,  Ans.        at  $3.63  per  perch,  cost 
$98.56.      112   ft.,   lineal 
measure,  at  $.25  per  foot,  cost  $28.     Therefore,  $98.56  +  $28= $126.56, 
the  cost  of  the  foundation. 

P  It  O  BIj  EM  S  . 

1.  How  much  will  it  cost  to  build  the  wall  of  a  cellar 
38  X  25  ft.,  7  ft.  deep,  and  18  in.  thick  ;  the  lower  4  ft.  to  be 
of  common  masonry,  at  $3.25  per  perch ;  the  next  3  ft.  to 
be  of  cut  stone,  at  15^  per  sq.  ft.  ?  Ans.  $155.97. 

2.  What  cost  the  walls  and  paving  of  a  conduit  250  ft. 
long,  4  ft.  wide,  and  2  ft.  high,  the  walls  being  built  of  cut 
stone  at  30)^  per  sq.  ft. ;  and  paving  laid  in  "  flags,"  at  25^ 
per  sq.  ft.  ? 

3.  What  cost  a  retaining  wall  built  of  rubble,  and  500  ft. 
long,  5  ft.  high,  and  2|^  ft.  thick,  at  $5.50  per  perch  ? 

Ans.  $1388.89. 

4.  A  stone  mason  received  $2777f  for  building  at  $5.50 
per  perch  a  stone  wall  500  ft.  long  and  5  ft.  high.  What 
was  the  thickness  of  the  wall  ?  Ans,  5  ft. 

5.  A  base  12  ft.  square  and  9  ft.  deep  was  built  for  a 
chimney  stack  at  the  rate  of  $3J  per  perch.  What  did  the 
base  cost  ? 

6.  The  cost  of  a  breakwater  100  ft.  long  and  20  ft.  thick, 
at  $1.75  per  perch,  cost  $2121.21.     How  high  was  it  built? 

Ans»  15  ft. 


APPLICATIONS.  257 

256.  Brick  Work. 
Brick   Work  is  usually  estimated  by  the  thousand 
hrickSf  though  sometimes  in  cubic  feet. 

IJ^  MEASURING   BRICKWORK, 

1.  Deduct  all  openings,  taking  tlie  7iet  vndtla.  on  the  outside  of  the 
wall,  and  running  up  to  the  spring  of  the  arch  in  circular  head 
openings. 

2.  Deduct  fireplaces  and  flues,  when  they  are  more  than  9  in.  square. 

3.  Ends  of  joists,  common  sized  lintels,  and  boxing  frames  are  not  to 
be  deducted. 

4.  Deduct  9  in.  square  flues  when  not  plastered  inside. 

When  bricks  are  2x4x8  in.,  it  requires  seven  bricks  to 
build  one  square  foot  of  a  wall  one  brick,  or  4J  inches  thick ; 
hence  it  will  take  14  bricks  to  build  a  square  foot  of  a  wall 
9  inches  thick  ;  21  bricks  to  build  a  square  foot  of  a  wall 
13  inches  thick,  &c. 

Therefore,  to  find  the  number  of  bricks  in  a  wall,  we  have 
this 

Rule. — Find  the  surface  of  the  wall  in  feet,  and  if  the 
wall  is  one  brick  thick,  multiply  the  number  of  feet  in  the 
surface  by  7 ;  if  two  bricks  thick,  by  14;  if  three  bricks 
thick,  by  21 ;  &c. 

Example  1. — How  many  bricks  in  a  wall  100  ft.  long,  8  ft. 

high,  and  13  in.  (3  bricks)  thick ;  and  what  will  the  wall 

cost,  at  $9  per  M.  for  the  brick,  and  $2  per  M.  for  laying 

the  same  ? 

SOLUTION.  Explanation. — The   number 

100  X  8  X  21  =  16800  of  Bq.ft.inthewall=800.    Since 

^9  X  16  800 ^151  2  *^®  y^^^  is  3  bricks  thick,  we 

49  s^  1A*«nn  —  ^  qq'«  multiply  by  21,  and  have  16800 

12  X  16.800  _  ^  33.6  ^^^^^    ^^^^^^  ^^  ^9  p^^  ^  ^^^ 

$184.8,  Ans.     $2  per  M.  for  laying  =--  $184.80. 
17 


258  DENOMINATE    NUMBERS. 

In  the  foregoing  Rule  and  Example,  we  suppose  the  brick  to  be 
2x4x8  in.,  and  in  estimating  the  sq,  ft.  of  the  surface,  we  suppose 
the  brick  to  be  laid  so  as  to  present  to  the  eye  its  3  x  8  in.  sides,  which 
gives  16  sq.  in.  surface.  Now,  7  of  these  surfaces  do  not  make  a 
sq.  ft.  ;  in  fact,  only  113  sq.  in. ;  but  if  we  estimate  the  mortar  as 
.438  in.  thick,  7  bricl?:s  will  present  a  surface  =  2  438  x  8.438  x  7  = 
144  sq.  in.  =  1  sq.  ft. 

If,  however,  as  is  often  the  case,  the  bricks  are  not  2x4x8  in.,  we 
must  exercise  our  judgment  in  estimating  the  number  of  bricks  used. 

Example  2. — How  many  pressed  bricks,  each  2^  by  4^  by 
8J  in.,  will  be  required  to  build  a  wall  40  ft.  long,  7  ft.  high, 
and  3  bricks  thick,  allowing  J  in.  between  bricks  for  mortar  ? 

SOLUTION. 

2^  +  1^  =  2|  in.,  thickness  of  brick  and  mortar. 
8J+^  =  8|  in.,  length  of  brick  and  mortar. 
8|  X  2i  =  21f  sq.  in.,  surface  of  brick  and  mortar. 
144-7-21|  =  6f?|,  number  of  bricks  in  1  square  foot. 
280  X  6fl^|  =  1843^,  number  bricks  in  wall,  1  brick  in  thickness. 
1843^x3  =  55291,       "  "      '•     "     3  bricks" 

Hence,  the  following 

Rule. — To  the  length  and  thickness  of  the  brick  in  inches, 
add  I  inch  ;  and  then  multiply  together  the  two  sums.  Divide 
144.  by  this  product  j  the  quotient  will  be  the  number  of  bricks 
in  a  square  foot  of  wall,  one  brick  thick.  Multiply  by  2,  3, 
or  Jf,  &c.,  to  find  the  number  of  bricks  in  a  wall  2, 3,  or  4,  Sc, 
bricks  in  thickness. 

In  estimating  brick  work  by  the  perch,  as  is  sometimes  done,  we 
proceed  as  in  measuring  Stone  Wokk,  except  that  the  isomers  a/re  not 
measured  tmce. 

PJt  OBTjEMS, 

1.  How  many  perches  in  an  ice  house  30  x  25  ft.,  and 
15  ft.  high,  the  walls  being  1|^  ft.  thick,  allowance  being 
made  for  1  door,  5  x  3  ft  ?  Ans.  93.63  +  P. 


APPLlCATlOirS.  259 

2.  At  $5  per  perch,  how  much  will  it  cost  to  build  a 
house  in  the  form  of  an  L,  the  main  building  being  40  x  25  ft., 
and  the  L  18  ft.  square  ;  walls  of  main  building  18  in.  thick, 
and  24  ft.  high,  and  those  of  the  L  9  in.  thick  and  14  ft. 
high ;  the  L  being  built  against  the  main  building ;  and 
two  doors,  each  7  x  3|-  ft.,  and  12  windows,  each  6x3  ft.,  in 
the  main  building,  and  1  door  G  x  3  ft.,  and  2  windows,  each 
5x3  ft,  in  the  L  to  be  deducted  ?  Ans.  $925.61. 

3.  What  will  it  cost  to  build  of  common  bricks  a  bam 
50  ft.  long,  25  ft.  wide,  and  20  ft.  high,  the  walls  being  13  in. 
thick,  when  brick  cost  $9  per  M.,  and  laying  the  brick  is 
paid  for  at  the  rate  of  $2  per  M.,  making  an  allowance  for 
2  doors,  each  3J  by  7,  1  door  8x7,  and  2  windows,  each  4 
by  4  ft?  Ans.  1641.333. 

Milwaukee  bricks  are  8^  x  4|^  x  3|  in. ;  Philadelphia  and  Baltimore, 
8Jx4ix2|. 

4.  How  many  Milwaukee  bricks  will  it  take  to  build  a 
house  45  x  35  ft,  25  ft.  high,  and  3  bricks  thick,  allowing 
for  3  doors,  each  8  x  3^  ft.,  and  24  windows,  each  7  x  4  ft.  ? 

Afis.  58996  bricks. 

5.  How  many  bricks  7 J  x  3|  x  2f  are  required  to  build  a 
square  enclosure  60  ft  on  each  side,  6  ft.  high  and  2  bricks 
thick,  allowance  being  made  for  a  gate  12  ft.  wide  ? 

Ans.  19169  bricks. 


257.  CAPACITIES. 

Example  1. — A  water  tank  built  in  a  rectangular  shape 
has  for  its  three  dimensions  5,  7  and  8  ft.  How  many 
gallons  will  it  hold  and  what  will  be  its  weight  ? 

NoTB. — A  cubie  foot  of  water  weighs  62.5  lb.  nearly. 


260  DENOMINATE   NUMBERS. 

Solution. — The  capacity  of  the  tank  in  cu.  ft.  is  5  x  7  x  8=280  cu.  ft. 
280  X  1728  =  483840,  the  number  of  cu.  in.  in  the  tank.  This  divided 
by  231,  the  number  of  cu.  in.  in  a  gallon  =  2094ff ,  the  number  of 
gallons.     62.5  x  280  =  17500,  weight  in  pounds. 

Example  2. — In  a  space  8  x  10  ft.  I  wish  to  build  a 
rectangular  tank  capable  of  holding  100  bbl.  How  high 
must  the  tank  be  ? 

Solution. — Since  there  are  81|  gallons  in  1  bbl.,  in  100  bbl  there 
are  3150  gal. ;  and  since  there  are  231  cu.  in,  in  1  gal.,  in  3150  gal. 
there  are  727650  cu.  in.,  which  is  the  capacity  of  the  tank.  Finally, 
since  the  bottom  of  the  tank  is  8  x  10  ft.  =  11520  sq.  in.,  the  depth  must 
be  as  many  inches  as  11520  is  contained  times  in  727650,  or  QB^y^^ 
times.  Therefore,  the  depth  must  be  Qd^is  iiiches  =  5  ft.  SyV^  in., 
Ans. 

JP  It  O  BL  EMS  . 

1.  How  many  bu.  bituminous  coal  may  be  carried  in  a 
barge  130  ft.  long,  25  ft.  wide,  and  7  ft.  deep  ? 

Ans.  130  X  25  X  7  X  A  (=  iHf )  =  1^625  bu. 

2.  How  many  bu.  of  bituminous  coal  can  be  stored  in  a 
shed  20  X 15  X 10  ft.  ?  Ans.  19284-  ^^• 

3.  A  teamster's  wagon-bed  is  12  ft.  x  ^  ft.  x  15  in. 
How  much  coal  does  it  hold  when  even  full  ?  Ans.  43  J^  bu. 

4.  The  same  teamster  raised  the  sides  and  ends  of  his 
Wagon-bed,  so  that  it  would  when  even  full,  hold  exactly  50  bu. 
coal.    How  much  did  he  increase  the  depth  ?  Ans.  2f  f  in. 

A  Ton  (2000  lb.) 

Of  loose  hay  on  scaffold  is  estimated  at  500  cu.  ft. 

Of  packed  hay  in  mow     "  "  '*  400  " 

Of  well-settled  stacked  hay  "  "  270  " 

Of  Lehigh  white  ash  coal  '•'  *'  34^  " 

Of  Schuylkill  white  ash  coal  "  '*  35  " 

Of  pink,  gray  and  red  ash  coal  *'  *'  36  " 

5.  How  many  lb.  of  hay  in  a  loose  heap,  13  x  9  x  5  ft.  ? 

Ans,  2340  lb. 


APPLICATIONS.  261 

6.  How  many  tons  of  hay  can  be  packed  in  a  mow 
20  X  20  X  6  ft.?  Ans.  6  tons. 

7.  How  many  tons  of  hay  in  a  well  settled  stack 
15  X  15  X  10  ft.  ?  A?is.  8|  tons. 

8.  My  coal  bin  is  9  x  7  x  6  ft.  How  many  tons  of  Lehigh 
white  ash  coal  will  fill  it  ?  A?is.  lOff  tons. 

9.  I  made  a  box  to  hold  1  T.  of  Schuylkill  white  ash  coal, 
and  occupy  only  a  space  2X3  ft.  on  the  floor.  How  high 
is  the  box?  Ans.  5  ft.  10  in. 

10.  A  teamster  wh^se  wagon-bed  was  7  ft.  long,  4J  ft 
wide,  and  12  in.  high,  claimed  that  when  the  bed  was  even 
full,  it  contained  just  1  T.  of  pink  ash  coal.  How  much 
did  it  lack  of  holding  that  quantity  ?  Ans.  ^  ton. 

11.  A  farmer  stored  away  300  bu.  of  rye,  150  bu.  wheat, 
70  bu.  barley,  and  350  bu.  shelled  corn.  What  must  have 
been  the  capacity  of  his  bins  ?  A  ns.  1082f  cu.  ft. 

12.  How  many  bu.  lime  can  I  burn  in  a  kiln  7  x  7  X 
14  ft.  ?  Ans.  441  bu. 

13.  Having  made  a  trench  15  ft.  long,  4  ft.  wide  and  3  ft. 
deep,  how  much  longer  must  I  make  it,  so  that  it  will  hold 
200  bu.  of  beets  ?  A7is.  10f|  ft.  longer. 

14.  How  many  cords  of  wood  can  be  placed  in  a  shed 
25  ft.  long,  24  ft.  wide,  and  16  ft.  high  ?  Aiis.  75  cd. 

15.  How  many  bu.  corn  in  the  ear,  in  a  "  crib  "  20  x  5  ft 
on  the  bottom,  20  x  8  ft.  on  the  top,  and  9  ft.  deep  ? 

In  some  localities,  2  heaped  bu.  of  corn  "in  the  ear"  are  called  1 
bu.,  which  would  make  our  Ans.  376^  bu.  In  other  localities,  how- 
ever, 2  eveji  bu.  of  '<ears"=l  bu.,  which  would  make  470^^  bu.  Ans. 

16.  If  a  freight  car  is  30  ft.  long  and  8  ft.  wide,  how  deep 
must  bulk  wheat  be,  in  order  that  the  car  may  contain 
24000  lb.,  the  greatest  weight  allowed  for  a  single  car  ? 

Ans,  24.888  in. +. 


26%  DEKOMINATE      NUMBERS. 

17.  How  deep  if  the  car  is  allowed  to  carry  only  10  T. 

Ans,  20.74    in.  +. 

18.  How  deep  if  the  grain  is  12  T.  rye,  56  lb.  to  the  bu.  ? 

Ans.  26.66  in.  +. 

19.  How  many  freight  cars  in  a  train  that  will  carry 
10000  bu.  wheat,  60  lb.  to  the  bu.,  each  car  carrying  its 
maximum  weight  ?  Ans,  2b  cslys. 

20.  How  many  bu.  shelled  corn  will  a  vessel  carry,  whose 
capacity  is  130  tons  ?  Ans. 

21.  A  farmer,  "closing  out  "  his  hay,  sold  what  he  had  in 
a  mow  18  x  20  ft.  and  20  ft.  deep,  at  the  rate  of  12  x  20  ft. 
and  1  ft.  deep  for  1  ton.  How  many  tons  less  than  he 
received  pay  for  did  the  farmer  sell  ?  Ans.  12  tons. 

TIME. 

258.   To  find  Longitude  from  Tinie,  and 
Time  from  Longitude, 

The  Meridian  of  any  place  is  a  north  and  south  line 
passing  through  that  place. 

The  Longitude  of  any  place  is  its  distance  reckoned 
in  degrees  from  some  assumed  meridian.  Thus,  if  a  place  is 
3°  east  of  another  it  is  said  to  have  3  degrees  of  longitude 
east  from  that  other  place.  Pittsburgh  is  nearly  3°  west  of 
Washington  and  Washington  is  nearly  3°  east  of  Pittsburgh. 

Longitude  may  be  reckoned  from  any  ineridian,  but  it  is 
customary  to  fix  upon  some  one  or  more  as  standards. 

The  English  reckon  Longitude  from  the  meridian  of 
Greenwich,  near  London ;  the  French  from  the  meridian  of 
Paris;  the  Eussians  from  the  meridian  of  St.  Peters- 
burgh,  &c. 


APPLICATIOlvrS.  2G3 

In  the  United  States,  longitude  is  reckoned  either  from 
the  meridian  of  WashingtoD,  or  from  that  of  Greenwich. 

The  earth  turns  from  west  to  east  one  rcYolution  in  24  hr. 
This  causes  the  Sun  to  appear  to  move  from  east  to  ivest 
around  the  earth  in  24  hr.  Hence,  the  Sun  appeal's  to  pass 
the  meridian  of  places  on  the  earth  in  order  from  east  to 
west  at  the  rate  of  ^  of  360°,  that  is,  15°,  every  hour  ;  and 
every  minute  of  time  he  appears  to  pass  through  -^  of  15°, 
which  is  -^  of  900',  or  15' ;  and  every  second  of  time,  he 
appears  to  pass  through  -^  of  15',  which  is  ^  of  900 ",  or 
15".    Hence  we  have  the 


TABLES. 


15° 

long.  =  1  hr.    time. 

1  hr.    time  =  15°  long. 

15' 

long.  =  1  min.  time. 

1  min.  time  =  15'  long. 

15' 

long.  =  1  sec.  time. 

1  sec.    time  =  15"  long. 

Example  1.  —  On  arriving  at  Washington  from  Pitts- 
burgh I  find  my  watch  12  min.  slow.  Supposing  my  watch 
to  keep  accurate  time,  how  do  I  account  for  the  apparent 
discrepancy  ? 

Solution. — Since  all  places  east  of  Pittsburgh  meet  the  Sun  earlier 
than  Pittsburgh  at  the  rate  of  1  hr.  for  every  15°,  I  conclude  that 
Washington  must  be  east  of  Pittsburgh,  and  since  it  meets  the  Sun 
12  minutes  earlier  than  Pittsburgh,  it  is  ^f  of  15°,  or  3°  east  of  Pitts- 
burgh. 

Example  2.  —  In  coming  from  London,  Eng.,  to  Wash- 
ington, D.  C,  I  find  that  my  "  Frodsham  "  watch  is  5  hr.  8 
min.  and  12.39  sec.  faster  than  Washington  time.  What 
does  this  fact  indicate  ? 


264  DEKOMIKATE     NUMBERS. 

SOLUTION.  Explanation. — Since  the  longitude  dif- 

hr      Tnin         sec  ^^^  ^*  *^®  ^^*®  °^  ^^°  ^°^  ®^^  ^^•'  ^^'  ^""^ 

k'        r.         -,  rt  qq         each  min.,  and  15"  for  each  sec,  by  multi- 
plying 5  hr.  8  min.  12.39  sec.  by  15,  we 


1^         obtain  the  difference  in  longitude  =  77°  3 


77°        3'  5  8"         5-^"-     -And  since  the  watch  was  fast,  we 

conclude  that  London  is  77°  3'  5.8"  east  of 
Washington.  Therefore,  in  finding  the  Longitude  from  Time,  we  use 
the  following 

KuLE. — Multiply  the  numher  of  hours,  minutes  and  sec- 
onds of  time  ly  15,  and  regard  the  product  as  degrees, 
minutes,  and  seconds  of  arc. 

Example  3.  —  When  it  is  11  o'clock,  A.  M.,  in  Wash- 
ington, D.  C,  what  time  is  it  at  London  ? 

SOLUTION  Explanation.  —  The  difference 

-  w  X  -,-,0        0/  K  on  of  Longitude  between  Washington 

^^UJ__ — t ^  and  London  is  77°  3'  5.8".    And 

6  hr.  8  min.   12.39  sec.     since  15°  of  long.  =  1  hr.  of  time, 
22  and  15'  of  long.  =  1  min.  of  time, 

and  15"  of  long.  =  1  sec.  of  time, 

it  is  evident  that  a  difference  of  77°  3'  5.8"  of  long,  must  equal  ^  of  as 
many  hr.,  min.,  and  sec.  of  time,  or  5  hr.  8  min.  12.39  sec.  And  since 
Washington  is  west  of  London,  it  will  meet  the  Sun  5  hr.  8  min.  12.39 
sec.  later  than  London  does.  Therefore,  when  it  is  11  o'clock,  A.  M., 
in  Washington,  it  is  11  o'clock,  A.  M.,  +  5  hr.  8  min.  12.39  sec,  or 
4  hr.  8  min.  12.39  sec.  P.  M.  in  London. 

Hence,  for  finding  the  Time  from  the  Longitude,  we  have  this 

EuLE. — Divide  the  degrees,  minutes,  and  seconds  of  arc 
by  15,  calling  the  quotients  respectively  hours,  minutes,  and 
seconds  of  time. 

PHOBIjEMS. 

1.  Cincinnati  time  is  29  min.  46.94  sec.  slower  than  that 
of  Washington.    What  is  the  longitude  of  Cincinnati  ? 

Ans.  r  26'  44.1",  W. 


APPLICATIONS.  2G5 

2.  Washington  time  is  5  hr.  17  min.  33  sec.  slower  than 
Paris  time.    How  many  degrees  is  Washington  west  of  Paris  ? 

A71S.  79°  23'  15". 

3.  How  far  is  Cincinnati  west  of  Paris  ?    Ans. 

4.  When  it  is  noon  at  Boston,  it  is  11  o'clock,  36  min. 
1.6  sec.  A.  M.  at  Washington.  What  is  the  longitude  of  Bos- 
ton? Ans.  5°  59'  36"  E.  from  Wash. 

5.  When  it  is  noon  at  Philadelphia,  it  is  11  o'c.  52  min. 
26.36  sec.  a.m.  at  Washington.  What  is  the  longitude  of 
Philadelphia  ?  Ans.  1°  53'  24.6"  E.  from  Wash. 

6.  When  it  is  noon  at  Washington,  it  is  12  o'c.  25  min. 
30  sec.  p.  M.  at  Santiago,  Chili.  What  is  the  longitude  of 
Santiago  ?  Ans.  6°  22'  30"  E.  from  Wash. 

7.  When  it  is  noon  at  New  Orleans,  it  is  6  hr.  10  sec.  at 
Greenwich.     What  is  the  longitude  of  New  Orleans  ? 

Ans.  90°  2'  30"  W.  from  Greenwich. 

8.  When  it  is  noon  at  Greenwich,  it  is  12  o'c.  50  min. 
18.66  sec.  P.M.  at  Copenhagen.  What  is  the  longitude  of 
Copenhagen  ?  Ans.  12°  34'  40"  E.  from  Greenwich. 

9.  If,  when  it  is  10  o'c.  A.  M.  at  Washington,  it  is  3  o'c. 
58  min.  8.8  sec.  p.  m.  at  Eome,  what  is  the  difference  of  lon- 
gitude between  Washington  and  Rome  ?    Ans.  89°  32'  12". 

Note, — In  subtracting  to  find  the  difference  of  time,  we  may  take 
10  lir.  from  (12  hr.  +  3  lir.  58  min.  8.8  sec.  =)  15  hr.  58  min.  8.8  sec, 
because  it  is  3  hr.  58  min.  8.8  sec.  more  than  12  o'c.  at  Rome. 

10.  Cincinnati  is  7°  26'  44.1"  W.  from  Washington.  How 
much  is  Cincinnati  time  behind  Washington  time  ? 

11.  The  beginning  of  an  eclipse  of  the  moon  occurred  in 
Washington  Mar.  10,  1876,  at  13  min.  6  sec.  after  12  A.  m. 
In  what  longitude  was  the  observer  who  saw  the  beginning 
of  the  same  eclipse  at  36  min.  5  sec.  after  12  A.  m.  ? 

A71S.  5°  44'  45"  E.  from  Wash. 


266  DENOMINATE     NUMBERS. 

12.  Paris  is  79°  23'  15"  E.  from  Washington.  How  much 
is  Paris  time  ahead  of  Washington  time  ? 

Ans.  5  hr.  17  min.  33  sec. 

13.  If  Albany  is  3°  18'  15"  E.  from  Washington,  how 
much  is  Albany  time  ahead  of  Washington  time  ? 

Ans.  13  min.  13  sec. 

14.  If  Boston  is  5°  59'  36"  E.  from  Washington,  what 
time  is  it  at  Washington,  when  it  is  noon  at  Boston  ? 

Ans.  11  o'c.  36  min.  1.6  sec.  a.  m. 

15.  If  Philadelphia  is  1°  53'  24.6"  E.  from  Washington, 
what  time  is  it  at  Washington,  when  it  is  noon  at  Phila- 
delphia ?  Ans.  11  o'c.  52  min.  26.36  sec.  a.  m. 

16.  If  Santiago,  Chili,  is  6°  22'  30"  E.  from  Washington, 
what  time  is  it  at  Santiago,  when  it  is  noon  at  Washing- 
ton ?  Ans.  25  min.  30  sec.  p.  m. 

17.  If  New  Orleans  is  90°  2'  30"  W.  from  Greenwich, 
what  time  is  it  at  Greenwich,  when  it  is  noon  at  New  Or- 
leans ?  Ans.  6  hr.  10  sec.  p.  m. 

18.  If  Copenhagen  is  12°  34'  40"  E.  from  Greenwich,  what 
time  is  it  at  Copenhagen,  when  it  is  noon  at  Greenwich  ? 

A71S.  50  min.  18.67  sec.  p.  m. 

19.  If  Rome  is  89°  32'  12"  E.  from  Washington,  what 
time  is  it  at  Washington,  when  it  is  3  o'c.  58  min.  8.8  sec. 
p.  M.  at  Rome  ?  A?is.  10  o'c.  a.  m. 

Note. — In  subtracting  the  difference  of  time  from  the  time  at  Rome, 
3  o'c.  may  be  called  the  15th  (12  o'c.  +  3  hr.)  hour  of  the  day. 

20.  An  observer  in  longitude  5°  44'  45"  E.  of  Washing- 
ton, noted  the  beginning  of  an  eclipse  of  the  moon  at  36 
min.  5  sec.  after  12  a.m.  on  the  10th  of  March,  1876. 
What  time  did  the  same  event  occur  in  Washington? 

'  Ans.  At  13  min.  6  sec.  a.  m..  Mar.  10,  1876. 


OUTLINE   OF   RATIO   AND   PROPORTION. 


r 

260.  Terms. 

r  261.  ^7i<€C€cien^. 
(  262.  Consequent. 

d 

H 

263.  First  Method, 
Numeration  and  Notation.  • 

264.  Second  MetliocL 

Principles. 

'  265.  First. 
'    266.  /SecoTMi. 

01 

.  267.  Third. 

Kinds. ' 

268.  /Simple. 

Direct.        ) 

I  271.  Variation. 
2 70.  Inverse.) 

^ 

.269.  Compound. 

o 

M 

H 
O 


273.  ^  Proportion, 

274,  Proportionals 


(275. 
(276. 


Terms. 


Three  Number$. 
Four  Numbers. 
'277.  Extremes. 

278.  Jfmn«. 

279.  Antecedents. 

280.  Consequents. 


281.   Principle. 

282.  /Sm^j^e.    284.  2?i^^e. 


Kinds 


.j 


283.  Compound.     285.  i^wfo. 


•i67 


CHAPTER    V. 


-^3 


KATiO'  jkmm  wmmwomwiom 


)<^i><$^?(i 


359.  HatiOf  in  Arithmetic,  is  the  relation  which  ono 
number  bears  to  another  of  the  same  kind. 

This  relation  is  found  by  dividing  one  of  the  numbers  by 
the  other.     The  quotient  is  also  the  value  of  the  ratio. 

The  divisor  may  be  regarded  as  the  measure  of  the  number  divided. 

Example  1. — What  is  the  relation  of  5  to  25  ? 

Ans.  5  is  -J-  of  25. 
Example  2.— What  is  the  relation  of  18  to  6  ? 

Ans.  18  is  3  times  6. 


In  Ex.  1,  we  notice  that  25  is  the  measure  of  5,  and—  is  the  value 
of  the  ratio. 

In  Ex.  3,  6  is  the  measure  of  18,  and  3  the  value  of  the  ratio. 

260.  Any  number  and  its  measure,  or  any  two  numbers 
that  are  compared,  are  the  Terms  of  a  ratio.  They  are 
also  a  couplet. 

The  Terms  are  named  from  the  order  of  their  position,  Antecedent 
(going  before),  and  Consequent  (following). 

261.  The  Antecedent  is  the  first  term,  or  the  numler 
to  he  divided,  or  measured. 

262.  The  Consequent  is  the  second  term,  or  the 
divisor,  or  7neasure. 


RATIO     AXD     PROPORTIOiq^. 


269 


Katio  is  indicated  in  two  ways,  viz. : 

363.  First. — By  a  fraction,  whose  numerator  is  the 
antecedent,  and  whose  denominator  is  the  consequent. 

264.  Second. — By  writing  the  consequent  after  the 
antecedent,  with  a  colon  between  them. 

Thus,  the  ratio  of  4  to  5  is  written  either  f ,  or  4  :  5,  and 
is  read,  "  four  is  to  five."  We  may  change  the  former  ex- 
pression into  the  latter,  or  tlie  reverse. 

The  terms  of  a  ratio  must  not  only  express  the  same  kind  of  quan- 
tity, but  also  the  same  denomination.  Thus,  the  ratio  of  3  quarts  to 
2  pecks  is  not  | ;  but  if  we  reduce  the  3  pk.  to  qt.,  the  ration  is  y\. 


Oral'OExfGBciXGX 


Example. — What  is  the  relation  of  4  to  5  ? 

Solution. — Since  we  find  the  relation  oi  one  number  to  another  by 
dividing  the  former  by  the  latter  (259),  the  yeiAtiou  of  4  to  5  is  f ; 
or,  4  -r-  5  ;  or,  4  :  5. 


PRO  li  L  E3tS. 

1.  What  is  the  relation  of  3  to  6  ?     5  to  8  ?     7  to  9  ? 

2.  What  is  the  -atio  of  15  to  3  ?     8  to  6  ?     12  to  9  ? 


3. 

11  to  12? 

7. 

6^  to  121  ? 

11. 

$20  to  $5  ? 

4. 

13  to  26  ? 

8. 

2.5  to  .5  ? 

12. 

£16  to  £4? 

5. 

.9  to  .27  ? 

9. 

100  to  .25? 

13. 

10s.  to  100s.  ? 

6. 

50  to  25  ? 

10. 

20  to  25  ? 

14. 

16d.  to  32d.  ? 

Since  the  ratio  of  two  numbers  is  expressed  by  a  fraction 
whose  numerator  is  the  antecedent,  and  whose  denominator 
is  the  consequent,  it  follows  that 


270  RATIO     AKD     PEOPORTION. 

265.  First. — Dividing  the  antecedent,  or  multiplying 
the  consequent,  divides  the  value  of  the  ratio  (163). 

266.  Secon"D. — Multiplying  the  antecedent  or  dividing 
the  consequent,  multiplies  the  ratio  (161). 

267.  Third. — Multiplying  or  dividing  both  terms  of  the 
ratio  by  the  same  number  does  not  affect  the  value  of  the 
ratio  (163). 

In  reference  to  the  number  of  its  terms,  a  ratio  is  either  Simple  or 
Compound. 

268.  A  Simple  Hatio  is  one  that  has  only  t^  o  terms ; 
as4:  5;  6:  8;  H;  ft;  &c. 

269.  A  Compound  Ratio  is  one  which  has  two  or 
more  pairs  of  terms.  Thus  |xj;  ^xfix4;  (2:4)x 
(7  :  8)  ;  (3  :  5)  X  (7  :  10)  x  (5  :  13)  are  compound  ratios. 


Example. — Reduce  3  :  5  and  10  :  13  to  a  simple  ratio. 

SOLUTION.  Explanation.— For 

3:5=4-)  2  convenience,   we    ex- 

-I Q  .  -I  Q  10  \  ^  ^  TF^^TT?  ^^  "  *  1'^'      press    the    ratios    in 

their  fractional  form 
and  proceed  as  in  multiplication  of  fractions,  obtaining  the  simple 
ratio  ^,  or  6  :  13.     Hence  the 

Rule. —  Write  the  given  ratios  in  their  fractional  form, 
and  multiply  them  together  as  in  multiplication  of  fractions  ; 
the  product  will  be  the  simple  ratio  required. 


BATIO     AND     PKOPOKTIOK. 


271 


P  J?  OBLEMS. 


(1) 


Reduce  the  following  to  simple  ratios 
(3)  (3) 


(4) 


3  :  15             2|  :     3 

20i  :     7 

7  :  41 

17  :  20            7:1 

5     :  11 

16  :  41 

16  :  19            5     :  16 

9     :  23 

9  :  54 

3: 

5        17  : 

15 

1 

Which  is  the  greater  ■{ 

7: 

8,  or  31  : 

42 

^? 

{ 

9: 

11           7  : 

35 

\ 

The  ratio  of  the  reciprocals  of  two  numbers  is  called  the  reciprocal, 
or  inverse,  ratio  of  those  numbers.  Thus,  the  reciprocal  of  2  is  |^; 
of  3, 1 ;  and  the  reciprocal  ratio  of  2  to  3  is  the  ratio  of  \  to  \.  The 
ratio  of  ^  to  i  is  ^  -H  1  =  I X  f  =  f ,  or  5  ;  2.    Hence, 

270.  The  Reciprocal  or  Inverse  Ratio  of  any  two  terms  is 
the  same  two  terms  with  their  positions  interchanged. 

2*71.  Variation. 

One  thing  varies  directly  as  another,  when  it  increases  as 
the  other  increases,  and  decreases  as  the  other  decreases. 
Thus,  if  the  rate  of  motion  be  uniform,  the  distances  moved 
over  will  vary  directly  as  the  time  ;  that  is,  in  twice  the 
time,  the  distance  moved  over  will  be  tivice  as  great,  and  in 
three  times  the  time,  it  will  be  three  times  as  great,  &c. 

One  thing  varies  inversely  as  another,  when  it  increases 
as  the  other  decreases,  and  decreases  as  the  other  increases. 
Thus,  the  time  of  moving  over  a  given  space  varies  inversely 
as  the  velocity  ;  that  is,  if  the  velocity  is  twice  as  great,  the 
space  will  be  moved  over  in  one-half  the  time ;  if  the  veloc- 
ity is  three  times  as  great  the  space  will  be  moved  over  in 
one-third  the  time;  &c. 


272  RATIO     AND      PROPORTIONS'. 

JPBOPOBTIOm 

273.  Proportion  is  equality  of  ratios. 

373.  ui.  proportion  is  a  statement  of  the  equality  of 
ratios.  Thus,  f  =  f ,  which  may  also  be  expressed  6:3  = 
8  :  4,  or  6  :  3  : :  8  :  4,  is  a  proportion.  The  last  is  read 
"  6  is  to  3  as  8  is  to  4."  A7iy  one  of  these  expressions  may 
he  changed  to  either  of  the  other  two  at  pleasure. 

274.  The  numbers  which  form  a  proportion  are  called 
proportionals,  or  tenns  of  the  proportion. 

275.  Three  numbers  are  proportionals  when  the  first  is 
to  the  second  as  the  second  is  to  the  third.  Thus,  2  :  4  ::  4  :  8. 
In  this  proportion  the  second  term,  4,  is  called  the  mean 
proportional  between  the  other  two  terms. 

276.  Four  numbers  are  proportionals  when  the  first  is  to 
the  second  as  the  third  is  to  the  fourth.    Thus,  2  :  4  : :  6  :  12. 

277.  The  Extremes  of  a  proportion  are  the  1st  and 
4th  terms. 

278.  The  Means  are  the  2d  and  3d  terms. 

279.  The  Antecedents  are  the  1st  and  3d  terms. 

280.  The  Consequents  are  the  2d  and  4th  terms. 

281.  In  every  proportion,  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means.  Thus,  in  the  proportions 
3  :  6  ::  5  :  10,  3x10  =  6x5;  and  7  :  2  ::  14  :  4,7x4 
=  14x2. 

This  Principle  is  called  the  test  of  the  accuracy  of  the 
proportion. 


RATIO     AND     PROPORTION. 


273 


Tell  which  of  the  followiDg  expressions  are  proportions: 


1.     3 

7  : 

:     5 

9. 

2.     6 

8  : 

:     3 

4. 

3.     7 

10  : 

:  14 

20. 

4.     6 

8  : 

:     5 

12. 

5. 

3  :  7 

:     4  :  9. 

6. 

2  :  1 

:     8  :  4. 

7. 

7  :  2 

:  21  :  6. 

8. 

5  :  3 

:  15  :  9. 

This  Principle  also  affords,  an  easy  method  of  finding  any 
one  term  that  may  be  wanting  in  a  proportion. 

Example  1.— What  is  the  4th  term  in  9  :  18  : :  5  :  (  )  ? 
SOLUTION.  Explanation. — The  product  of  the  means 

18  X  5  _  .^     .  ^'^x  18  =  90. 

Q        —  1^>  -a-US.         The  product  of  the  extremes  must  also  be 
90  (281).     Since  one  of  the  extremes  is  9, 
the  other  extreme  must  be  90  h-  9  =  10,  Ana. 

Example  2. — What  is  the  2d  term  in  3 


SOLUTION. 

3  X  12 


—  9,  Ans. 


(    )  ::  4  :  12? 
Explanation. — The  product  of  the  ex- 
tremes is  3  X  12  =  36. 

The  product  of  the  means  must  also  be 
36  (281).     Since  one  of  the  means  is  4,  the 
other  mean  must  be  36  -5-  4  =  9,  A71S. 

KuLE  I. — Divide  the  product  of  the  means  hy  the  given 
extreme;  the  quotient  is  the  other  extreme. 

II.  Divide  the  product  of  the  extremes  ly  the  given  mean  ; 
the  quotient  is  the  other  mean. 

Cancelling  common  factors  from  the  antecedents  or  consequents,  or 
either  couplet,  wiU  not  destroy  the  proportion;  because  the  results  will 
in  each  case  stand  the  test. 

rnO  BLEM  s. 

Find  the  unknown  term  in  the  following : 

7.  £(     )  :  £12  : :  15  :  3. 

8.  12  :  $5  : :  20  lb.  :  (     )  lb. 

9.  8;^  :  24^  ::  (     )  qt.  :  6qt. 

10.  lOpwt. :  (  )pwt.  ::  15  :  6. 

11.  (   )  oz.  :  12  oz.  : :  $5  :  |6. 

12.  7T. :  11 T.::  14001b. :  (  ). 


1. 

1 

3::4:(    ). 

2. 

4  • 

2::10:(     ). 

3. 

6 

8  : :  (     )  :  12. 

4 

9  : 

5  ::  (     )  :  10. 

5. 

3 

(     )::8:6. 

6. 

5 

:  (    )  ::20:12. 

274  RATIO     AK1>     PROPOETION. 

In  reference  to  the  number  of  terms  in  their  ratios,  proportions  are 
Simple  or  Compound. 

282.  A  Simple  JProportion  is  a  statement  of  the 
equality  of  two  simple  ratios.  Thus,  5  :  8  : :  10  :  16,  or 
A  =z  ^^,  is  a  simple  proportion. 

283.  A  Compound  I^roportion  is  a  statement  of 

the  equality   of  a  simple  and  a  compound  ratio,  or  of 

2  •  3 
two  compound  ratios.     Thus,  g  ;  g  : :  4  :  6,  or  2  x  6  : 

3  X  9  : :  4  :  6,  or =  f,  represents  the  equality  of  a 

compound   and    simple   ratio;    while  q  !  g  -^   o    •   4'  ^^ 

1  X  3  :  2  X  6  ::  5  X  2  :  10  X  4,  or  ^-^  =  T^r^,  rep- 

2x6        10  X  4       ^ 

resents  the  equality  of  two  compound  ratios. 


SIMPLE   PROPORTION. 

Simple  Proportion  is  employed  to  find  a  fourth  propor- 
tional, when  three  are  given. 


0raL  3E:Coi^ci^G^ 


•-♦•• 

Example. — If  9  lb.  sugar  cost  81)^,  what  cost  12  lb.  ? 

Solution.— If  9  lb.  cost  81;^.,  12  lb.  will  cost  more  in  the  ratio  of 
12  to  9,  or  y-  =  f  as  much. 

I  of  81/  =  108/.  Therefore,  if  9  lb.  of  sugar  cost  81/,  12  lb.  cost 
108/,  or  $1.08. 

Pn  OBLEMS. 

1.  If  6  yd.  carpet  cost  $15,  what  do  10  yd.  cost? 

2.  If  a  tree  20  ft.  high  casts  a  shadow  8  ft.  long,  how 
high  is  the  tree  whose  shadow  is  32  ft.  long  ? 


RATIO     AND     PROPORTIOI^.  275 

3.  7  bbl.  molasses  cost  $63.     What  cost  14  bbl.  ? 

4.  2  lb.  coffee  last  6  persons  8  days.  How  long  will  it 
last  24  persons  ? 

5.  Paid  $1.20  for  3  lb.  butter.  How  many  lb.  can  be 
bought  for  $3.60  ? 

6.  5  horses  eat  a  ton  of  hay  in  8  months.  How  long 
would  it  take  2  horses  to  eat  the  same  amount  ? 

7.  If  the  boarding  of  3  men  for  a  week  costs  $21,  how 
many  men  can  be  boarded  the  same  length  of  time  for 
$63  ? 

8.  If  a  10^  loaf  of  bread  weighs  12  oz.  when  flour  is  $8 
a  bbl.,  how  much  should  it  weigh  when  flour  is  $16  a  bbl.? 

9.  If  3  lb.  sugar  cost  33^  what  will  6  lb.  cost  ? 

10.  If  4  bu.  apples  cost  $3,  what  do  8  bu.  cost  ? 

11.  If  5  T.  iron  cost  $400,  what  will  8  T.  cost? 

12.  If  6  bu.  wheat  cost  $9,  what  do  5  bu.  cost  ? 

13.  If  11  lb.  coffee  cost  $1.65,  what  do  12  lb.  cost? 

14.  If  12  lb.  rice  cost  72)^,  what  do  11  lb.  cost  ? 

15.  If  15  bbl.  beef  contain  3000  lb.,  how  many  do  20  bbl. 

contain  ? 

.> _ 


t'Wr^tten  ^:^erciXeXt 


Example  1. — If  50  bu.  wheat  cost  $130,  what  will  60  bu. 
cost  at  the  same  rate  ? 

Explanation. — Since  both  terms  of 
$  a  ratio  must  be  like  numbers,  we  see  that 

(156)  we  must  compare  50  6?y.  with  60  Mi.;  and 
$130  with  the  requisite  number  of  dol- 
lars in  the  result. 

Since  proportion  is  an  equality  of  ra- 

50  )  7800  tios,  the  ratio  of  50  to  60,  or  |^,  must  be 

$156.  Ans.  equal  to  the  ratio  of  $130  to  $(  ),  or  p^ ; 


276  RATIO     AND     PROPORTION. 

or  50  :  60  :  :  $130  :  $(    ).     Solving  by  281,  Rule  I.,  we  have  f^  x 
$130  =  $156. 

Example  2. — If  10  men  can  build  a  boat  in  12  days,  how 

long  will  it  take  15  men  to  do  the  same  work  ? 

SOLUTION.  Explanation.— If  10  men  build  the 

men    men     days    days  ^^^^  ^  ^^  days,  15  men  will  build  it  in 

15  :  10  : :   12  :   (8)  less  time  ;  in  other  words,  the  relation 

12  X  10  ,        ,  of  15  to  10,  or  If ,  determines  the  relation 

~"Y5        —  ^  ^^'  ^^^^'  of  13  da.  to  the  required  time. 

384.  EuLE. — MaJce  that  number  luMcli  is  of  the  same 
hind  as  the  required  answer,  the  third  term. 

Then  detertnine  from  the  nature  of  the  question  whether 
the  ansiver  is  to  be  greater,  or  less  thari  the  third  term. 

If  greater,  turite  the  less  of  the  reinaining  two  terms  as  the 
first  term  of  the  proportion,  and  the  greater  as  the  second. 
If  less,  ivrite  the  greater  of  the  two  remaining  terms  as  the 
first  and  the  less  as  the  second. 

Find  the  fourth  term  by  2S1,  Rule  I. 

As  the  processes  are  all  multiplication  and  division,  it  is  evident 
that  the  work  may  frequently  be  shortened  by  cancellation. 

This  and  all  similar  examples  may  be  solved  by  Analysis.     Thus, 

Solution  1. — If  10  men  can  build  a  boat  in  12  da.,  it  would  take 
1  man  10  times  12  da.,  or  120  da.  And  if  1  man  will  build  the  boat 
in  120  da.,  15  men  will  build  it  in  -^^  of  120  da.,  or  8  da.,  Ans. 

2.  If  10  men  build  a  boat  in  13  da.,  15  men  will  build  it  in  the  same 
part  of  13  da.  that  10  is  of  15,  or  i§  =  |  of  13  da.,  or  8  da..  Ana. 

mo  B  LJiJM  s. 

1.  If  50  bu.  wheat  cost  $75,  what  cost  40  bii.?    Ans.  $60. 

2.  If  20  yd.  cloth  cost  $40,  how  much  do  18  yd.  cost  ? 

A?is.  $33. 

3.  If  12  lb.  tea  cost  $10,  how  much  would  25  lb.  cost  ? 

Ans.  $20f . 
4  If  7  caps  cost  $9,  what  do  10  cost  ?  A7is.  $12f 


RATIO     AND     PROPORTIOK.  277 

5.  If  $63  buy  21  hats,  how  many  will  $48  buy  ? 

Ans.  16  hats. 

6.  If  30  bu.  wheat  make  6  bbl.  flour,  how  many  bu.  will 
be  needed  to  make  8  bbl.?  Ans.  40  bu. 

7.  If  50  bu.  wheat  make  10  bbl.  flour,  how  many  bbl.  of 
flour  will  60  bu.  make  ?  Ans.  12  bbl. 

8.  If  a  man  walks  60  mi.  in  4  da.,  how  far  would  he  walk 
in  6  da.,  at  the  same  rate?  Ans.  90  mi. 

9.  If  a  man  travels  132  mi.  in  6  da.,  how  long  will  it  take 
him  to  travel  462  mi.,  at  the  same  rate  ?  Ans.  21  da. 

10.  If  a  man,  at  the  rate  of  3  mi.  an  hr.,  walks  a  certain 
distance  in  12  hr.,  how  long  would  it  take  him  to  walk  the 
same  distance,  at  6  mi.  an  hr.  ?  Ans.  6  hr. 

11.  If  a  ship's  provisions  will  last  36  men  216  da.,  how 
long  would  they  last  60  men  ?  A7is.  129.6  da. 

12.  If  24  men  can  build  a  wall  in  42  da.,  how  long  would 
it  take  16  men  ?  •  A7is.  63  da. 

13.  If  36  men  can  build  a  wall  in  42  da.,  how  many  men 
would  build  it  in  63  da.  ?  Ans.  24  men. 

14.  If  a  ship's  provisions  would  last  48  men  162  da. ,  how 
many  men  would  they  last  324  da.?  Ans.  24  men. 

15.  If  40  gal.  syrup  cost  £6  10s.  (130s.),  what  cost  50  gal.  ? 

Ans.  £8  2s.  6d. 

16.  If  5  lb.  8  oz.  tea  cost  $8.80,  how  much  do  3  lb.  2  oz. 
cost?  Ans.  $5. 

17.  If  5  bu.  4  qt.  wheat  cost  $12.30,  how  much  do  7  bu. 
3pk.  cost?  A?is.  $18.60. 

18.  If  a  ship  has  water  enough  to  allow  each  of  her  crew 
of  50  men  H  pt.  per  day,  till  the  end  of  her  voyage,  how 
much  can  be  allowed  if  she  takes  on  board  30  more  men  ? 

Ans.  3}  gi. 


278  EATIO     AND     PROPOETIOl^. 

19.  If  $150  gain  $12,  what  sum  will  gain  $1200,  in  the 
same  time,  at  the  same  rate  ?  Ans.  $15000. 

Explanation. — Al- 

SOLUTION.  though  all  the  terms 

gain         gain        capital         capital  are  dollars,  the  third 

$12    :    $1200  :  :  $150    :    $  (15000)  term  is  a  different /a;id 

1200  X   150 


12 


from  the   1st   and  2d 
=  $15000,^^25.       terms.     Therefore,  we 
state  the  Problem  as 


in  the  solution,  comparing  gain  with  gain,  and  capital  with  capital. 

20.  How  long  will  oats  last  12  horses,  which  would  last 
4  horses  24  wk.  ?  Ans.  8  wk. 

21.  How  long  will  hay  last  15  cows,  which  would  last 
8  cows  20  wk.  ?  Ajis.  10|  wk. 

22.  How  many  men  can  do  that  work  in  15  da.,  which 
would  take  25  men  90  da.  ?  Ans.  150  men. 

23.  If  5  lb.  sugar  cost  80^,  what  cost  |  lb.  ?     Ans. 

24.  If  i  lb.  tea  cost  37^^,  what  cost  5  lb.  ?    A7is.  $7.50. 

25.  If  6  T.  iron  cost  £90  18s.  6d.,  what  cost  ^  T.  ? 

26.  If  -J  bbl.  flour  costs  $||,  what  costs  |  bbl.  ? 

A71S.  $6.75. 

27.  An  upright  stick  1  ft.  6  in.  high,  casts  a  shadow  3  ft., 
when  a  tree  casts  a  shadow  192J  ft.  How  high  was  the 
tree?  Ans.  96|  ft 

28.  If  I  borrow  $150  for  6  mo.,  how  long  should  the 
lender  have  $300  of  my  money  to  equalize  favors  ? 

Ans.  3  mo. 

29.  If  I  borrow  $150  for  6  mo,,  how  many  dollars  should 
the  lender  have  5  mo.  to  equalize  favors?  Ans.  $180. 

30.  If  a  set  of  men  can  lay  the  brick  of  a  building  in 
18  days,  working  10  hr.  per  day,  in  how  many  days  could 
fchey  lay  them  working  14  hr.  per  day?  Ans. 


SOLUTION. 

sq.  rd.    sq.  rd.          rd. 

rd. 

4    :     20     ::     1     : 

(5) 

RATIO     AND     PROPORTION^.  279 

31.  A  rectangular  lot  containing  20  sq.  rd.  is  4  rd.  wide. 

How  long  is  it  ? 

Explanation.  —  (See  Figure, 
209.)    Four  sq.   rd.  along  one 
side  are  1  rod  wide.     Therefore, 
4  sq.  rd.  are  to  the  20  sq.  rd.,  as 
the  width  of  the  4  sq.  rd.  is  to  width  of  the  20  sq.  rd. 

32.  If  2|  mo.  boarding  cost  $50f,  what  would  15^^  mo. 
boarding  cost,  at  the  same  rate  ?  Ans.  $314.65. 

33.  A  lot  40  rd.  long  and  4  rd.  wide  contains  an  acre. 
How  long  must  an  acre  lot  be,  if  8  rd.  wide  ?    Ans.  20  rd. 

34.  If  sound  goes  6160  ft.  in  5|  sec,  how  far  does  it  go  in 
16i  sec.  ?  Ans.  18480  ft. 

35.  If  the  moon  moves  on  the  average  52°  42'  20"  in  4  da., 
in  what  time  would  it  make  a  complete  circuit  of  the  earth  ? 

Ans.  27.32  da. +  . 

36.  If  a  clock  loses  1  min.  45  sec.  in  36  hr.,  how  much 
time  will  it  lose  in  2  wk.  ?  Ans,  16  min.  20  sec. 

37.  If  $175  gain  $21,  what  do  $100  gain?       Ans.  $12. 

38.  If  $350  lose  $42,  what  do  $250  lose  ?        Ans.  $30. 

39.  If  $800  pay  $3  tax,  what  does  $1  pay  ?  Ans.  $.003}. 

40.  If  $2500  earn  $1000,  what  do  $1000  earn  ? 

Ans.  $400. 

41.  What  does  $1  earn  ?  Ans.  $.40. 

42.  If  the  13^  loaf  weigh  15  oz.  when  flour  is  $8  per  bbl., 
what  should  it  weigh  at  $12  per  bbl.  ?  Ans.  10  oz. 

43.  What  should  it  weigh  at  $6  per  bbl.?     Ans.  20  oz. 

44.  A  40  gal.  barrel  is  filled  with  a  defective  gallon 
measure,  and  thus  appears  to  hold  42|  gal.  How  much 
short  was  the  gallon  measure  ?  Ans.  2  gi. 

45.  If  I  of  a  barrel  of  cider  cost  $1^,  what  is  the  cost 
of  I  of  a  barrel  ?  Ans.  $lfi-. 


280  RATIO     AJ^D     PROPORTION. 

Compound  Proportion, 

Compound  Proportion  is  used  when  the  solution 
of  a  problem  depends  upon  two  or  more  pairs  of  conditions. 


-^i 


Otal  ^xfoBci^eX 


Example.— If  $50  gain  $2  in  8  mo.,  how  much  will  $25 
gain  in  10  mo.  ? 

Solution.— If  $50  gain  $2  in  8  mo. ,  $25  will  in  tlie  same  time  gain 
less,  in  the  ratio  of  25  to  50,  or  ff  =  | ;  |  of  $2  =  $1. 

If  $25  gain  $1  in  8  mo,,  in  10  mo.  it  will  gain  more  in  the  ratio  of 
10  to  8,  or  Jfl  =  f  ;  f  of  $1  =  $li. 

Therefore,  if  $50  gain  $2  in  8  mo.,  $25  will  gain  $1|^  in  10  mo. 

pn  obIjEMS. 

1.  If  8  men  mow  32  A.  in  3  da.,  how  many  men  can  mow 
48  A.  in  12  da.  ? 

2.  If  10  loaves  are  sufficient  for  20  men  1  da.,  how  many 
loaves  will  supply  40  men  2  da.  ? 

3.  If  5  horses  eat  10  tons  of  hay  in  12  mo.,  how  many  tons 
will  8  horses  consume  in  6  mo.  ? 

4.  If  3  boys  earn  $7  in  4  da.,  how  much  can  9  boys  earn 
in  20  da.? 

5.  If  400  bu.  coal  are  required  to  supply  8  fires  5  mo.,  how 
much  coal  would  2  fires  consume  in  10  mo.  ? 

6.  If  5  weeks'  boarding  of  6  men  cost  $150,  what  will 
7  wk.  boarding  of  5  men  cost  ? 

7.  If  10  men  can  build  a  wall  30  ft.  long  in  5  da.,  how 
long  will  it  take  16  men  to  build  a  similar  wall  96  ft.  long? 


BATIO     AND     PROPORTIOiq-.  281 

8.  If  25  oranges  are  worth  50  lemons,  and  6  lemons  are 
worth  24^,  how  much  are  15  oranges  worth  ? 

9.  If  10  men,  by  working  10  hr.  per  da.,  can  build  a  boat 
in  8  da.,  how  many  da.  would  it  take  12  men  to  do  the  work, 
if  they  wrought  only  8  hr.  per  da.  ? 

10.  If  a  5^  loaf  weighs  10  oz.  when  flour  is  18  a  bbl.,  how 
much  should  an  8^  loaf  weigh,  when  flour  is  $5  a  bbl.  ? 


t^Wrdtten  ^]^er  ci^esTHr 


Example. — If  16  men  in  20  da.  10  hr.  long  can  build 
5  mi.  of  railroad,  in  how  many  da.  12  hr.  long  can  64  men 
build  30  miles  of  railroad  ? 

SOLUTION. 

64  men  :    16  men 

12  hr.     :    10  hr.    ::    20  da.    :    (    )  da. 
5  mi.     :    30  mi. 

64  X  12  X  5    :    16  X  10  X  30   ::    20   :    (25) 

Explanation. — We  write  20  da.  for  3d  tenn,  because  our  answer, 
or  4th  term,  is  required  in  da. 

Since  64  men  will  perform  the  work  in  less  time  than  16  men,  our 
4th  term  must  be  less  than  the  3d,  and  we  therefore  write  64  men  for 
our  1st  term  and  16  men  for  the  2d. 

Since  it  would  take  a  shorter  time  to  do  a  piece  of  work  by  working 
12  hr.  per  day  than  by  working  10  hr.  per  day,  our  4th  term  must  be 
less  than  the  3d,  and  we  therefore  write  12  hr.  for  the  1st  term  and 
10  hr.  for  the  2d 

Since  it  would  take  more  time  to  build  30  mi.  of  road  than  5,  the 
4th  must  be  greater  than  the  3d  term,  and  we  therefore  write  5  mi. 
for  the  1st  and  30  mi.  for  the  2d  term. 

We  then  reduce  the  compound  ratio  to  a  simple  one  (269),  and 
find  the  4th  term  by  Rule  (284),  or  by  (281,  Rule  I.) 


282 


RATIO     AND     PEOPORTION. 


285.  Rule. —  Write  as  the  tJiird  term  that  number  which 
is  of  the  same  kind  as  the  required  answer. 

Then,  of  the  other  terms,  form  a  compound  ratio,  being 
careful  to  write  the  terms  in  pairs,  and  to  state  each  pair  of 
terms,  as  if  the  value  of  the  4th  term  depended  on  that  pair 
alone. 

After  which,  find  the  fourth  term  in  accordance  with  the 
Principle  of  Proportion  (281). 

This  and  all  similar  examples  may  be  solved  by  Analysis. 

Thus, 

Solution. — If  16  men  can  build  5  mi.  (or  any  other  numb.er  of 
miles)  of  railroad  in  20  da.,  64  men  will  build  the  same  amount  in  the 
same  part  of  20  da.  that  16  is  of  64,  that  is  ^  of  20  da.,  or  5  da. ;  and  if 
they  can  build  5  mi.  in  5  da.,  they  can  build  30  miles  in  f  of  30  da.  = 
30  da.  And  if  they  can  build  it  in  30  da.,  by  working  10  hr.  per  da., 
they  can,  by  working  12  hr.  per  day,  build  it  in  f|,  or  f  of  30  da.,  or 
25  da.,  Atis. 

This  and  similar  examples  may  also  be  solved  by  Cause 
and  Effect.    Thus, 

SOLUTION. 

BECOND  CAUSE. 

r       64  men      1 

,  j       working       I 

'  I  12  hr.  per  da.,  [ 

J      I     for(    )da.     J 

We  have  now  to  find  the  number  of  days  in  the  Sec.  cause.  And 
since  this  is  a  proportion,  the  product  of  the  extremes  equals  the 
product  of  the  means,  and  we  therefore  can  readily  fill  the  blank  (  ) 
by  Rule  I,  or  II.  (281). 

PiJ  OB  LEMS. 

1.  If  6  men  build  a  wall  3  rd.  long  in  5  da.,  how  many 
men  would  build  a  wall  15  rd.  long  in  10  da.  ?    Ans. 

2.  If  24  cows  eat  36  bu.  com  meal  in  12  da.,  how  many 
bu.  will  last  20  oows  60  da.  ?  Ans.  150  bu. 


FIRST  CAITSB. 

r       1(5  men 

J       working 
I  lOhr.perda., 
[     for  20  da. 


EFFECT  OF 

EFFECT  OF 

1st  Cause.  ] 

Zd  Cause. 

5  mi.      1  . 

30mi. 

of         1  • 

of 

railroad  - 

.  railroad. 

RATIO     AND     PPtOPORTION.  283 

3.  If  675  gain  $12  in  2  yr.,  what  sum  will  gain  $150  in 
18  mo.  ?  Ans.  $1250. 

4.  If  16  men  earn  $640  in  26  da.,  what  will  48  men  earn 
in  104  da.  ?  Ans.  $7680. 

5.  If  3  men  in  18  da.  earn  $162,  how  many  dollars  will 
9  men  earn  in  12  da.,  with  like  wages?  A7is.  $324. 

6.  If  10  persons  spend  $600  in  8  mo.,  how  much  would 
8  persons  spend  in  12  mo.,  at  the  same  rate?    Ans.  $720. 

7.  If  7  persons  spend  $273  in  2  mo.,  in  how  many  mo. 
would  5  persons  spend  $650?  Ans.  6f  mo. 

8.  If  $250  gain  $15  in  1  yr.,  in  how  many  months  will 
$750  gain  $37.50  ?  Ans.  10  mo. 

9.  If  $300  gain  $9  in  3  mo.,  what  wiH  gain  $50  in  10  mo. 

Ans.  $500. 

10.  If  8  men  dig  a  ditch  6  rd.  long  in  10  da.,  how  many 
men  would  dig  27  rd.  long  in  12  da.  ?  Ans.  30  men. 

11.  If  8  men  dig  a  ditch  6  rd.  long  in  10  da.,  in  what 
length  of  time  would  30  men  dig  27  rd.  long?    Ans.  12  da. 

12.  If  8  men  dig  a  ditch  6  rd.  long  in  10  da.,  how  far 
would  30  men  dig  in  12  da.  ?  Ans.  27  rd. 

13.  If  12  horses  eat  18  bu.  of  oats  in  6  da.,  how  many 
bu.  would  last  20  horses  30  da.?  Ans.  150  bu. 

14.  If  6  horses  eat  4^  bu.  oats  in  3  da.,  how  many  days 
will  100  bu.  last  10  horses?  A7is.  40  da. 

15.  If  4  horses  eat  12  bu.  of  oats  in  6  da.,  how  many 
horses  will  require  75  bu.  in  30  da.  ?  A71S.  5  horses. 

16.  If  11  men  mow  45  A.  of  grass  in  6  da.  of  10  hr.  each, 
how  many  men  will  be  required  to  mow  81  A.  in  12  da.  of 
11  hr.  each  ?  Ans.  9  men. 

17.  If  11  men  mow  45  A.  in  6  da.  of  10  hr.  each,  how 
many  hr.  per  da.  must  9  men  mow,  to  finish  81  A.  in  12  da.? 

Ans. 


284 


RATIO     AND     PROPORTION 


18.  If  11  men  mow  45  A.  in  6  da.  of  10  hr.  each,  how 
many  A.  will  9  men  mow  in  12  da.  of  11  hr.  each  ? 

Ans.  81  A. 

19.  If  108  men  build  a  fort  in  24 J  da.  of  12|-  hr.  each,  in 
how  many  da.  would  84  men  build  it,  working  10^  hr.  per  da.  ? 

Ans.  37^  da. 
Distributive  JPropoi^tion, 
286.    Distributive   Proportion  is,  as  its  name 
implies,  proportion  applied  to  the  distribution  of  a  quantity 
into  parts  which  have  a  given  ratio.     It  is  sometimes  called 
Partitive  Proportion. 


f  ©pat^xfGitciXeX  -^ 


-•'¥•- 


Example.— What  parts  of  20  are  as  3  to  1. 

Solution. — Since  20  is  to  be  divided  into  parts  which  are  as  3  to  1, 
it  must  be  divided  into  3  +  1,  or  4  equal  parts,  and  3  and  1  of  these 
parts  taken.  20  -f-  4  =  5,  or  1  part ;  three  of  these  parts  is  15,  and 
one  of  them  is  5.  Therefore,  the  parts  of  20  which  are  as  3  to  1,  are 
15  and  5. 

PMO  BLE  MS. 


What  parts 


1.  Of  30  are  as  2  to  3  ? 

2.  Of  63  are  as  4  to  5  ? 

3.  Of  96  are  as  3  to  5  ? 

4.  Of  84  are  as  4  to  2  ? 


5.  Of   21  are  as  2  to  1  ? 

6.  Of   49  are  as  2  to  7  ? 

7.  Of  105  are  as  7  to  8? 

8.  Of  110  are  as  10  to  11? 


9.  Anna  and  Mary,  aged  respectively,  5  and  7  yr.,  found 
$1.20,  and  divided  it  in  the  ratio  of  their  ages.  How  much 
did  each  receive  ? 

10.  Jane,  Ann,  and  Eliza  contributed  to  the  purchase  of 
a  present  of  an  $18  clock  to  their  teacher,  in  the  ratio  of  1, 
2,  and  3.    How  much  did  each  contribute? 


EATIO     AND     PROPORTION. 


285 


11.  E  and  F  spent  $29,  F  spending  6^  times  as  much  as 
E.    How  much  did  each  spend  ? 

12.  Frank  and  James  have  75^  ;  Frank  has  J  as  much  as 
James.     How  much  has  each  ? 

13.  Fifty  has  two  parts,  one  of  which  is  ^  the  other. 
What  are  the  parts  ? 


t  Wi'itten  ^?(ferci«e^-f 


Example. — Resolve  75  into  two  parts,  of  which  the  first 


shall  be  to  the  second  as  7  to  8. 

SOLUTION. 

7   :     8  ,  ratio  of  parts. 
7  +    8  =  15,  no.  of  parts. 
75  -^  15  =    5,  one  of  the  parts. 
5  X    7  =  35,  1st  Part, 
5  X    8  =  40,  2d  Part, 


i\ 


Ans. 


Explanation.  —  From  the 
conditions  of  the  Example,  75 
is  to  be  divided  in  the  ratio  of 
7  to  8,  or  in  such  a  manner  that 
the  sum  of  the  same  equimul- 
tiple of  7  and  of  8  shall  be 
equal  to  75.  And  since  75  is 
to  equal  an  equimultiple  of 
7  and  8  at  the  same  time,  it  must  be  an  equimultiple  of  their  sum, 
15.  Now  75  contains  15,  5  times  ;  therefore,  one  part  of  75  must  con- 
tain 7,  5  times,  and  the  other  part  must  contain  8,  5  times.  Hence  the 
parts  are  5  x  7  =  35,  and  5  x  8  =  40. 

SOLUTION  BY  PROPORTION. 

7  4-8  =  15,-15:  7::  75:  (35)  1st  Part. 
7  +  8  =  15,  —  15 :  8  : :  75 :  (40 )  2d  Part. 
387.   Rule. — Divide  the  given  number  by  the  sum  of  the 
terms  of  the  ratio,  and  multiply  the  quotient  by  each  term. 
Or,  As  the  sum.  of  the  terms  of  the  ratio  is  to  one  of  them, 
so  is  the  number  to  be  resolved,  to  its  corresponding  part. 


PR  OBLJEMS 


What  parts 

1.  Of48areasl,  2,  and3? 

2.  Of  108  are  as  2,  3,  and  4  ? 


Ans.  8,  16,  and  24. 
Ans.  24,  36,  and  48. 


286  RATIO     AND     PROPORTIOif. 

3.  Of  108  are  as  5,  6,  and  7  ?  Ans.  30,  36,  and  42. 

4.  Of  64  are  as  1,  3,  5,  and.7  ?     Ans.  4,  12,  20,  and  28. 

5.  Of  120  are  as  H,  2J,  3^,  and  ^  ? 

Ans.  15,  25,  35,  and  45. 

6.  Of  95  are  as  3J-,  4J,  5^-,  and  6J? 

^?i5.  16J,  21J-,  26J,  and  31J. 

7.  A  and  B  engaged  in  a  speculation.  A  invested  $300, 
and  B  $500.  They  gained  $80.  What  was  each  man's 
share  of  the  gain  ?  Ans.  A%  $30 ;  B's,  $50. 

8.  A,  B,  and  0  bought  a  farm  of  650  A.,  for  which  A  paid 
$2000;  B,  $3000;  and  0,  $1500.  How  many  A.  should 
each  have  ?  Ans.  A,  200  A. ;  B,  300  A. ;  0,  150  A. 

9.  Four  men.  A,  B,  C,  and  D,  held  a  pasture  for  $54.  A 
put  in  3  cows ;  B  put  in  5 ;  0  put  in  4 ;  and  D  put  in  6. 
How  much  should  each  pay  ? 

A71S.  A,  $9;  B,  115;  0,  $12  ;  D,  $18. 

10.  Three  farmers  bought  a  threshing  machine  in  part- 
nership for  $330.  The  first  gave  $80;  the  second,  $100; 
and  the  third,  $150.  What  part  of  the  time  should  each 
have  the  use  of  it  ? 

Ans.  The  1st,  -^;  the  2d,  -J^;  and  the  3d,  J|- 
of  the  time. 

11.  Eesolve  84  into  two  such  parts  that  one  shall  be 

6  times  the  other. 

Solution. — Since  84  is  to  be  the  same  equimultiple  of  1  and  6,  it 
must  be  the  equimuhiple  of  their  sum,  7.  If  84  is  a  multiple  of  7, 
the  factor  by  which  7  is  multiplied  must  be  |  of  84,  or  12.  Therefore, 
the  numbers  are  12  times  1,  or  12 ;  and  12  times  6,  or  72. 

12.  Divide  40  into  two  parts,  so  that  one  shall  be  3  times 
the  other. 

13.  Divide  144  into  two  such  parts  that  one  shall  be  eleven 
times  the  other. 


RATIO     AND     PROPORTION^.  28? 

14.  A  and  B  divide  960  A.  so  that  A  has  5  times  as  many 
A.  as  B.     How  many  has  each  ? 

Ans.  A,  800;  and  B,  160  A. 

15.  M  and  N.  together  have  $6300 ;  and  M  has  1  J-  times 
as  much  as  N.     How  much  has  each  ? 

Ans.  M,  $3600;  N,  $2700. 

16.  Kesolve  100  into  two  parts,  one  of  which  shall  be  16 
less  than  the  other. 

SOLUTION. 

100  —  16  =  84 
84  -^  (1  +  1  =)  2  =  42,  Less  Number.        ) 

42  +  13  =  58,  Greater  Number.  )         ' 

Explanation. — Since  the  less  and  the  less  plus  16  =  100,  16  plus 
twice  the  less  =  100 ;  and  twice  the  less  =  16  less  than  100,  or  84. 
Hence  the  less  =  |  of  84,  or  42  ;  and  the  greater,  42  +  16  =  58. 

17.  Eesolve  150  into  two  parts,  one  of  which  is  40  less 
than  the  other.  Ans.  55  and  95. 

18.  Kesolve  75  into  two  parts,  one  of  which  is  15  more 
than  the  other.  Ans.  30  and  45. 

19.  H  has  $1500  more  than  I ;  both  have  $25000.     How 
much  has  each  ?  Ans.  H,  $13250  ;  I,  $11750. 

20.  M  is  21  yr.  younger  than  0 ;  both  ages  equal  85  yr. 
How  old  is  each  ?  Ans.  M,  32  yr. ;  0,  53  yr. 

21.  What  part  of  400  is  100  more  than  twice  the  other 
part  ?  Ans.  300.     The  other  part  100. 

22.  Two  men  built  2695  rods  of  fence,  and  the  first  built 
I  as  much  as  the  second ;  how  many  rods  did  each  build  ? 

Ans.  1st.  1225 ;  2d.  1470. 

23.  Divide  15400  into  four  parts  that  shall  be  to  each 
other  as  i,  -J,  i,  -^.  Ans.  6000,  4000,  3000,  2400. 

24.  Resolve  132  into  two  such  parts,  that  4  times  the  first 
equals  7  times  the  second. 


SOLUTION. 

4.  X  1st  = 

7  X  2d; 

1  X  1st  = 

i  X  2d. 

J: 

K^d  -\-l  X  2d  = 

-.{i^l)x2d  = 

¥ 

X  2d 

=  132. 

i 

X  2d 

=  12;  and 

i 

X  2d 

=  48,  the  second  p^rt. 

i 

X  ^^ 

=  7  X  12  : 

=  84,  the  first  part. 

288  RATIO     AND     PROPORTION. 

Explanation. — If 
4  times  the  1st  =  7 
times  the  2d,  then  the 
1st  =  :|^  of  7  times  the 
2d,  or  I  of  the  2d. 
Hence,  |  of  the  2d  +  f 
of  the  2d,  or  -^^i  of  the 
2d  =  132.  Therefore, 
i  of  the  2d  =  ^  of 
132,  or  12 ;  and  |  of 
the  2d  =  4  X  12,  or  48  ; 
and  I  of  the  2d  =  7  x 
12,  or  84,  the  first. 

25.  Eesolve  96  into  two  parts,  of  which  3  times  one  equals 
5  times  the  other.  Aiis.  60  and  36. 

26.  Eesolve  720  into  two  parts,  of  which  2  times  one  equals 
7  times  the  other.  Ans.  560  and  160. 

27.  Resolve  65  into  two  such  parts  that  f  of  one  shall  equal 
4"  of  the  other. 

SOLUTION.  EXPLANA- 

%  X  lst  =  4t  X  2d;  i  X  lst  =  ^  X  2d;         tion.-H  f  of 

^Xlst  =  ^x  2d.  ^^^  ^^^""^l'  ^ 

of  the  other 
fx^^  +  ^X^^=(f  +  ^-^)x^^  =  65;     theniofth; 

and^-  x2d  =  ^  =  5;  ^  x2d=l!  x5  =  35;     1st  =  ^  of  f, 
and  f  X  ^^  =  6  X  5  =  30,  1st  part.  or  f  of  the 

2d,  and  the 
whole  of  the  1st  =  3  times  f ,  or  f  of  the  2d  Hence,  f  +  f ,  or  i^  of 
the  2d  =  both  parts,  or  65 ;  and  -^  of  65,  or  5,  is  |  of  the  2d,  and  7 
times  5  =  35,  is  the  2d  part.     And  65  —  35  =  30,  is  the  1st  part. 

28.  Resolve  50  into  two  such  parts  that  -J-  of  one  equals  J 
of  the  other.  Ans.  20  and  30. 

29.  Resolve  25  into  two  such  parts  that  twice  one  equals 
J  of  the  other.  Ans.  5  and  20. 

30.  If  }  of  A's  age  is  f  of  B's,  and  the  sum  of  their  ages  is 
51  yr.,  what  are  their  ages  ?     Ans.  A's,  24  yr. ;  B's,  27  yr. 

31.  C  and  D  own  400  A.,  and  |  of  C's  farm  equals  f  of 
D's.     How  many  A.  has  each  ? 

Ans.  C,  160  ;  and  D,  240  A. 


OUTLINE    OF    PERCENTAGE. 


Ph 


BEBC^Ifl^fJ^J^.-i^QO.  2d  Meaning. 


T'TT'TJlVr^    i  291.  RATE  PER. CENT, 

U 


2*er  Cenf. 
292.  RATE. 
294.  AMOUNT. 
295.  D INFERENCE. 

296.  NUMERATIOJ^  AND  NOTATION. 

^  *  ^^^    i  297.  /.    .  .  298.  RULES.     299.  II,    .  300.  RULES. 
CASES,  j  301.  iZ/. .  302.  RULES.     303.  iF.  .  304.  RULES. 

305.  FORMULAE. 

^,TKTT,M^c   ^  308.  Co»«.  309.  Rate  %. 

TERMS.]  3jQ^  Profit  and  Loss.  311.  Selling  Trice, 
312.  -T.   .    .    .   313.  Rules, 
314.  JX.     .   .   315.  Rules, 
31G,  111.  .    .   317.  7;»t/e*. 
318,  IF.     .    .    319.  Rules. 


©OH 


CASES. 


H      TERMS.   )  3J3J3^  Policy, 

Q   I  (327.  Underwriters, 


0<^KINDS. 


15 

O 

H 

O 


CASES. 


323.  J'/rR.  334.  Marine. 

325.  Health.        336.  ii/e. 

t  339.  /. 
328.  J^ire  anrf  Marine. ■{  330.  /Z 

(33i./zr. 

332.  I>i/e  jfn^Mrance, 


r334.  Agent. 

335.  Commission  Merchant. 

336.  Broker. 
TEli::^S.'{  337.  Brokerage. 

I  338.  Consignment, 

339.  Consignor. 
*^340,  Consignee. 
CASES    i  341.  Z.  342.  TJ. 

TER3IS. 


347.  JLwtoMn*. 


KINDS. 


346.  Principal. 
35V.  Rate  %. 

As  to  ntode  of  computation.  {HI-  f^^ 

As  to  Rate  %  charged. 

r352.  I. 


350.  Legal. 

351.  lUegaL 


CASES 


•  '<'< 

CC*-'cJ 

r 


354.  II. 
.  \  356.  J/r. 


19 


.   353.  Rules. 

.   355.  Rules. 

.   357.  Rules. 

]  358.  JF.    .    .   359.  JJu/e*. 

1360.  V.  .   .   .   361.  JJt«?c*. 

<-  3  6  3 .  Days  of  Grace.        364.  MatuHty. 
rr  ««»,„      -^65.  The ''Promise.'"      368.  i^a<!e. 
lerms.  |  3^9  Maker  or  Drawer.  370.  Holder  or  Payee. 
'371.  Indorser.  372.  /Vo^i. 

q-r^    P«r««7      (374.  U.S.  Rule. 

p2S2n«  1  3'''^-  Commercial  Rule. 
payment.  \^^q  Connecticut  Rule. 

366.  Negotiable. 

367.  Jbin^. 

377.  Compound  Interest.   .   .   .   378.  Sule. 
L379.  raftfe.  .  .  .  380.  Rule. 

989 


Kinds. 


OUTLINE-Continued. 


o 


o 

M 

H 

O 

S 
<   ^ 


% 


f  382.  Trade. 


383.  Bunk. 


984.  A  Bank, 


nep06tt.     1  388.  A  Chuck. 
389.  ^  Bank  of  Issue. 
1^  '^  39(^.  ^  ^anA;  of  Discount. 

391.  Trwe,  or  Ef/uitable 392.  Comparison 

of  True  and  Hank. 
BANK  ^.  999.  Case  I.   .  994:. Bule.  39 5. Case  II.   99G.  JRule. 
BIS.    ■{  397.  Case  III.  998.  Bule.  399.  Case  IV.  400.  Bute. 
COUNT.  (4:01.  Case  V.  .  40^. Bule. 

TRUE   {409.  Case!   .  404.  Bule.  40  5.  Case  11.   406.  Bule. 
BIS-    \  407.  Case  IIL  408.  BuU.  409.  Cawe  lY.  410.  iJw^c 
COf/JVT.  (411.  Ca««  F.  .  412.i?wfe. 

415.  .4  Charter. 
Certificates.       418.  Stockholders. 
I  419.  ^<  Par.        ( 
Value.-l  420.  ylftowe  Par.-^  422.  Market  Value. 

(  421.  ^ctow  Par.  { 
423.  Bividends.  424.  Assessments. 

425.  Bonds.  426.  Computations. 

•H  raoa    4  n^^.^ft  S  42,9.  Inland  Draft. 
i^g  I  *^^'      -^'^"^*-  i  430.  J^or^^n  M/  0/  Exchange. 
H^  j  431.  J?afe  ofTSxchange.         432.  Course  of  Exchange. 

435  .  Bays  of  Grace. 


xti 

it 


^414.  ^  Corporation. 
'416.  Shares.        417 


^•^  I  433.  >St<//it  Braft. 
Sffi      434.  Time  Braft.  .   . 
^   ^436.  Principle. 


05 

444. 

i<i 

, 

C 

S 

as 

.*  tj 

^ 

"^  ^ 

;s 

(5 

438.  Birect.  J  440.  mi  ^ 

( 441.  Property,  -j  ^^3  pergonal. 

439.  Indirect. 
Birect  Taxes. 


Terms. 


ElKDS. 


Deduc- 
tions. 


{ 


Rule. 

447.  Customs. 

449.  TaHff. 

451.  Cuswm  Bouses. 


445 
446.  Duties. 
448.  JSScasg. 
450.  Bevenue. 
45J8.  Invoice. 

453.  Specific. 

454.  .4cf  Valorem. 

455.  Breakage.         f  4  5  9.   Gross 

456.  Leakage.  J  Weight. 

457.  T'are.  1  460.   Net 
458  Z>ra/«  or  7Ve«.  L  Weight. 

(463.  I*artners.  463.  Company. 

<  464.  Capital.  465.  Bividend. 

{  466.  Assessment.      467.  Liabilities. 
zi««  (  469.  l>c6<or«.        4H 9.  Creditors. 

BANKRUPTCY.!  '*''^*   ^«»e/«.  472.  J\et  i>rocced» 


461. 
PARTNERSHIP. " 


(  473.  Assignee. 
f  475.  Equated  Time.  476.  J^oeai  JDafe. 
477.  Term  0/  Credit.  478^  verage  Term  of  Credit. 
KINBS.— 479.  Simple.    480.  Compound. 

481.  AN  ACCOUNT.\lll   ^'^^^^ce. 

COlif  PUT  A- {Simple  Average 484.  Rule. 

TIONS.      S  Compound  Average  .  .  486.  Rule, 


CHAPTER    VI. 

SECTION    I. 


wmmmM.mTA(^m 


-^. 


^^. 


288.  I^er  cent  (contraction  for  per  centum)  means, 
by  the  hundred.  Thus,  5  per  cent  of  $1  is  yfo  of  a  dollar, 
or  5  cents. 

The  term  percentaffe,  as  commonly  used,  has  two 
meanings. 

389.  1.  It  is  the  name  of  a  process  of  computing  by 
hundredths.  Thus,  finding  the  value  of  5  per  cent  of  a 
dollar,  is  working  an  example  in  percentage. 

390.  2.  It  is  a  name  applied  to  any  number  of  hun- 
dredths of  fif?iy^7u'?2^  that  is  regarded  as  a  basis  of  computation. 
Thus,  in  the  example  5  per  cent  of  $1  =  5c.,  5c.  is  the 
percentage,  $1  the  basis,  or  hase. 

Per  cent  is  frequently  used  in  the  same  sense  as  the 
second  meaning  of  Percentage. 

291.  Hate  per  Cent  is  the  number  of  hundredths. 
Thus,  in  the  expression  6  per  cent  of  $1,  0  is  tlie  rate  per 
cent.  In  computations,  we  give  the  rate  per  cent  a  denom- 
inator 100,  and  thus  obtain  a  common  or  decimal  fraction 
called 

292.  The  Mate,  an  expression  denoting  what  part  of 

291 


292 


PERCEI^T  AGE. 


the  base  equals  the  percentage.  Thus,  in  6  per  cent  of  $1  = 
6c.,  the  rate  is  -^^s,  -^,  or  .06. 

293.  The  Base  of  percentage  is  that  number  ot 
which  a  certain  per  cent  is  computed. 

394.  The  Amount  is  the  sum  of  the  base  and  per- 
centage. 

295.  The  Difference,  or  Remainder ,  is  the  Base 
less  the  Percentage. 

296.  Bate  per  cent  is  usually  written  with  the  sign  %, 

Thus  5  per  cent  may  be  written  b%.     The  sign  %,  is  always 

read  **  per  cent."    Thus,  6%  is  read  "  Six  per  cent." 

Note. — The  sign  is  generally  used  in  business,  but  in  calculations 
rate  only  is  used,  and  is  written  as  a  decimal  or  common  fraction. 

Thus,  tlie  rate  per  cent  Q\  is  the  rate  — ^,  or  yV>  or  .06^,  or  .0625  ;  the 

rate  %  \%\  is  the  rate  --^,  or  \,  or  .12^,  or  .125. 

One  iwr  cent  of  a  number  is  ^^  of  that  numher  ;  less  than 
one  per  cent  of  a  numher  is  less  than  y^  of  that  number. 
One  hundred  per  cent  of  a  number  is  that  number,  because 
it  is  {^  of  the  number.  Tioo  hundred  per  cent  of  a  number 
is  twice  that  number,  or  f f§-  of  it ;  etc. 


.01 
.02 
.03 

4 

100 
_5_ 
100 

6 
100 


EXERCISES. 

Read  as  per  cent  the  following  rates : 

.oii 

.105 

.75 

.OOJ 

.02} 

.125 

1.75 

.005 

.07^ 

.161 

2.00 

.0075 

9i 
100 

87i 
100 

350 
100 

100 

10 

25 

425 

i 

100 

100 

100 

100 

lOi 

50 

505 

125 

100 

100 

100 

100 

.001 

2.1 

2. 
200 
100 
300 
100 
600 
100 


PERCENTAGE. 

2 

Write  as  rates  the 

following  rates  per 

cent: 

>Seveii  per  cent. 

100  per 

cent. 

75  per  cent 

Eight     "      " 

1    " 

a 

750    «      " 

Nine       "      " 

2     " 

a 

3     "      « 

Ten        "      « 

200     " 

a 

33^  "      « 

Twenty "      « 

50     " 

a 

1    a         « 

Thirty-five    " 

500     " 

u 

6i  "      « 

Sixty 

i" 

6( 

125     "      " 

Eighty-five    " 

n " 

(( 

12J  "      « 

293 


Table  showing  what  ^jar^  of  anything  a  certain  7'ate  %  o/ 
it  equals : 


1^  =  TiTr- 

6,^    =  fV. 

H%     =   A- 

16f;?  =  f 

3$?=  W- 

6i^  -  A- 

9^      =  tItt- 

25^    =i. 

3^  =  T^- 

6|;^  =  T^- 

10^    =  A. 

33i$^  =  i- 

4$^=  1^- 

^%     =TiF- 

13!^    =  A- 

66|^  =  |. 

5^=  A. 

8^    -   A- 

13i$g=    i- 

K%    =i. 

SECTION    II. 
GENERAL  CASES  OF  BEHCEj^TAGE. 

CASE    I. 

297.  J5ase  ancf  l^a^e  jjer  cent  given,  to  find 
Percentage* 


©ral^ 


■^♦•- 


ExAMPLE  1.-5^  of  $10  is  what  part  of  it  ?     10^  ? 
Solution.— 5  fc  of  $10  is  yf^,  or  i^  of  it.     10%  of  $10  is  ^^VV  or 


of  it. 


294 


PERCENTAGE, 


Example  2.— How  much  is  5%  of  110?    10^? 
Solution.— 5%  of  $iO  is  j^^,  or  ^V  of  $10  =  50;^.    10%  of  $10  ia 
i^oV  or  j\  of  $10  =  $1. 


Find 

1.     1%  of  $100. 

5. 

2.    5%  of  60  bu. 

6. 

3.    Sfo  of  125. 

7. 

4.  25^  of  £64. 

8. 

Find 
9.  100^  of  3  cows. 

10.  150%  of  100  A. 

11.  250%  of  20  yd. 

12.  40%  of  1000  gr. 


I*JiOBIj:EMS, 

Find 

7%  of  $100. 
i%  of  1200. 
2i%  of  40  yd. 
8.  37i^  of  32  mi. 

13.  A  merchant  gained  in  one  year  28%  of  his  capital. 
What  part  of  his  capital  equals  his  gain  ? 

14.  A  man  whose  farm  contained  280  acres,  bought  as 
many  acres  more  as  equaled  10%  of  what  he  already  had. 
How  many  acres  did  he  buy  ? 

15.  In  a  school  of  300  pupils,  40%  are  girls  and  the  re- 
mainder are  boys.     How  many  of  each  ? 

16.  A  gentleman  bequeathed  20%  of  his  estate  to  a  college, 
10%  to  an  orphan  asylum,  and  10%  to  a  public  library,  and 
the  remainder  he  gave  to  his  children.  What  %  of  his 
estate  did  the  children  receive  ? 

17.  Suppose  there  were  three,  how  many  %  would  that  be 
for  each  ;  and  what  part  of  the  estate  would  each  receive  ? 

18.  Suppose  the  estate  worth  $200000;  how  much  would 
each  child  receive  ? 


bWii{iim^:?Ce^ci^e^b 


Example. — Find  15%  of  30  bu.  of  wheat. 

1st  solution.  Explanation.  — 15%  of  30  ba.  is  the 

30  bu.  Base.  same  as  .15  of  30  bu,  and  .15  of  30  bu.  is 

.15  Rate.  4.5  bu.     Therefore,  15%  of  30  bu.  is  4.5 

bu. 


4.5  bu.  Percentage. 


PERCEKTAQE.  295 

2d  solution.  Explanation.— Since  15%  means  15 

15^  _    16_  _  _3_  hundredths,  we  take  that  part  of  30  bu. 

100%  ~  100  "~  20  that  15%  is  of  100 /c,  that  is,  ^fp^,  or  ^% 

/o  of  30  bu.  =  ^  bu.  «^  20  bu.  =  4|  bu.  Ans. 

Note.  —  We  might  solve  the  Example  by  Proportion,  thus : 
100%  :  15%.  :  :  30  bu.  :  (4|-)  bu. 

298.   Rules. — I.  Multiply  the  Base  hy  the  Rate. 
II.  Take  that  part  of  the  Base  that  the  given  rate  %  is  of 
200%. 

PROBTjEMS, 

1.  Find  25%  of  64  horses.  Ans.  16  horses. 

2.  Find  35%  of  $125.  Ans.  143.75. 

3.  Find  45%  of  £13^.  Ans.  £5.88. 

4.  Find  16J%  of  10  A.  Ans.  1.65  A. 

5.  Find  21%  of  18.5  bu.  Ans.  3.885  bu. 

6.  Find  2|%  of  17.25  mi.  Ans.  .474375  mi. 

7.  Find  \%  of  $2.25.  Ans.  $.0028125. 

8.  Find  1%  of  7.5s.  Ans.  3.15  far. 

9.  Find  3^%  of  $51.  Ans.  $1.70. 

10.  Find  12J%  of  a  ship.  Ans.  ^  of  a  ship. 

11.  Find  16|^  of  an  oil-well.  Ans.  -J-  of  an  oil-well. 

12.  Find  33J%  of  an  iron-mill.         Ans.  ^  of  an  iron-mill. 

13.  Find  133^%  of  $100.  Ans.  $133.33f 

14.  The  distance  from  Pittsburgh  to  Philadelphia  is  355.3 
mi.,  and  from  Pittsburgh  to  New  York  is  25%  more.  What 
is  the  distance  from  Pittsburgh  to  New  York  ? 

Ans.  4:4A  mi. 

15.  The  distance  from  New  York  to  Pittsburgh  is  444  mi., 
and  the  distance  from  Philadelphia  to  Pittsburgh  is  20% 
less.    What  is  the  distance  between  the  two  latter  places  ? 

Ans.  355.2  mi. 


296 


PERCENTAGE. 


16.  A  merchant  sold  goods  on  Monday  amounting  to 
$275.37J;  and  on  Tuesday,  sold  37^%  more.  What  was  the 
amount  of  Tuesday's  sales  ?  Ans.  ^378.643^. 

17.  At  Pittsburgh  the  spring  rain-fall  averages  9.38  in. ; 
in  summer  its  average  is  4.7^  more.  What  is  the  summer 
rain-fall  ?  Ans.  9.82  in. 

18.  The  average  cost  of  each  pupil  in  the  common  schools 
of  Pennsylvania  in  1869  was  $.97  per  month ;  in  1849,  it 
was  49.4^  less.     What  was  the  cost  in  1849  ?     Ans.  $.49. 

19.  The  average  length  of  the  school  term  in  1863  was 
5.433  mo.;   in   1869   it  was   11.2^  more.     What  was  the 


increase 


Ans.  .608  mo. 


CASE    TI. 
299.  Base  and  Percentage  given  to  find  the 
Hate  JPer  Cent. 


€)i*ar^xfoiici<GX 


^ 


r 


Example  1.— What  %  of  $10  is  ^  of  it  ? 

Solution.— The  whole  of  $10  is  100^  of  it.    Hence,  ^  of  $10 
must  be  -^  of  100%  of  it,  or  5%. 

Example  2. — Six  dollars  is  what  %  of  $60  ? 

Solution.— Six  dollars  is  ^  of  $60 ;  $60  is  100%  of  itself.    There- 
fore, $6  must  be  yV  of  100%  of  $60,  or  10%  of  $60. 

PRO  BT^  EMS, 


1.  2  is  1  ? 

2.  4  is  2  ? 

3.  5  is  2  ? 

4.  4  is  1  ? 

5.  3isi? 


What 

6.  $8  is  $6  ? 

7.  $9  is  $2.25  ? 

8.  10^  is  12f  ? 

9.  15  fr.  is  12  fr.  ? 
10.  8  dr.  is  5  dr.? 


of 

11.  16  horses  is  4  horses  ? 

12.  25  men  is  5  men  ? 

13.  30  pens  is  6  pens  ? 

14.  50  books  is  100  books  ? 

15.  64  hoes  is  8  hoes? 


PEECEKTAGE. 


297 


IG.  A  farmer  raised  100  busTiels  of  wheat  and  150  bushels 
of  rye.     What  %  more  rye  than  wheat  did  he  have  ? 

17.  A  farmer  sold  \  of  his  wheat  to  one  man  and  ^  to 
another.     What  %  had  he  remaining  ? 

18.  Having  drawD  out  7  gallons  from  a  barrel  of  vinegar, 
I  find  there  28  gallons  remaining.  What  %  of  the  whole 
did  I  draw  ? 

19.  In  a  school  of  300  pupils,  120  are  girls.  What  %  of 
the  school  equals  the  number  of  boys  ? 

20.  A  farmer  who  owned  40  head  of  cattle  purchased  8 
more.     What  ^  did  he  add  to  his  stock  ? 

21.  What  %  of  his  former  stock  equals  his  present  stock  ? 

22.  A  newsboy  lost  12  out  of  36  papers  which  he  had 
bought.     What  per  cent  was  his  loss  ? 

23.  A  clerk  whose  salary  is  $40  per  month,  spends  $5  for 
cigars,  $10  for  boarding,  $6  for  car-fare,  and  $3  for  knick- 
knacks.     What  %  of  his  salary  has  he  left  ? 

,^, 


+1Wi  itten  ^^vferci^es^^ 


— »« . 

Example. — What  %  of  25  marks  are  7  marks  ? 

Explanation. — Since  the  percentage  is 
the  product  of  the  base  and  rate,  used  as 
factors,  the  rate  is  the  quotient  obtained 
by  dividing  the  percentage  by  the  base : 
and  7  divided  by  25  is  .28  =  28%. 

Explanation.— 7  is  ^\  of  25.  But,  25 
is  the  Base  =  100%.  Therefore,  7  marks 
must  be  ^^  of  100%  =28%. 


1st  solution. 

7  ^  25  =  .28  =:  28^. 

2d  solution. 

A  of  100^  =z  28^, 
or, 
25:  7::  100^:  (28)^. 


300.  ^VLE,— Divide  the  Percentage  hy  the  Base.    Or, 
Tahe  such  a  part  of  100%  as  the  Percentage  is  of  the  Base. 


298 


PERCENTAGE, 


PJt  OB  IjEM8» 

What  per  cent 


1.  Of  $12  is  19  ?     Ans,  75^. 

2.  Of  £16J  is  £8}  ?  ^^^s.  50^. 

3.  Of  87^^  is  12i.^'  ?  ^?i5.  l^%. 

4.  Of20fr.  islifr.? 

Ans.  1\%. 


5.  Of  6  bu.  is  li  bu.  ?  J:?i5. 25^. 

6.  Of  16  pk.  is  1  pk.  ?  Ans.  6^%. 

7.  Of£lis5d.?       Ans.  %^%. 

8.  Of  12.5  mi.  is  .75  mi.? 

Ans.  6%. 


9.  A  farmer  raised  500  bushels  of  rye.  He  sold  A,  150  bu. ; 
B,  160  bu. ;  0,  50  bu.    What  %  had  he  remaining  ? 

Ans.  28j^. 

10.  After  selling  ^  and  i  of  my  mill,  how  many  %  do  I 
own  ?  Ans.  33^^. 

11.  Paid  $4.50  for  the  use  of  $50.     What  %  did  I  pay  ? 

Ans.  9%. 

12.  The  average  length  of  the  school  term  in  Pa.,  in  1863, 
RTas  5.433  mo. ;  in  1869  it  was  6.04  mo.  What  was  the  rate 
per  cent  of  increase  ?     What  was  the  rate  of  increase  ? 

Ans.  n.2%',  rate,  .112. 

13.  At  Pittsburgh  the  spring  rain-fall  is  9.38  in. ;  in  sum- 
mer, 9.82  in.  How  many  %  of  the  former  is  the  latter,  and 
how  many  %  of  the  latter  is  the  former? 

Ans.  104.7^+;  95.5^  +  . 

CASE    III. 
301.  JPercentage  and  Hate  Per  Cent  given 
to  find  Base, 

•>• 


Qhal  ^Xoitci^Gj^ 


••♦>• ■ — 

Example. — 5  bu.  are  20^  of  how  many  bu.  ? 

SoLtJTiON. — 20%  of  any  number  =  ^  of  that  number. 
6  bu.  must  be  ^  of  5  times  5  bu.,  or  25  bu. 


Therefore, 


PERCENTAGE. 


299 


1.  Is  10^  10^  ? 

2.  Is  20  da.,  15%? 

3.  Is  2i  hr.,  b%  ? 


7.  Is  8  fur.,  3%  ? 

8.  Is  3  mi.,  33^%  ? 

9.  Is  7i  far.,  15%  ? 


Of  what 

4.  Is  10  cii.,  2%  ? 

5.  Is  £5,  10%  ? 

6.  Is  8s.,  5%  ? 

10.  James  had  18  marbles,  which  was  25%  of  what  Thomas 
had.     How  many  had  Thomas? 

11.  15  men,  which  was  15%  of  a  company,  deserted.    What 
was  the  number  of  men  in  this  company? 

12.  Out  of  $75,  a  lady  gave  20%  for  a  hat.    Wh^t  had  she 
left? 


T  ^r^tteii  iE:?^erci5e^t 


EXAMPLE.- 


IST  SOLUTION. 

25  bu.  -j-  .05  =  500  bu 


25  X 
25 


25  bu.  are  5%  of  how  many  bu.  ? 

Explanation. — Since  the  percentage  is 
the  product  of  the  base  and  rate,  used  as 
factors,  the  base  is  the  quotient  obtained 
by  dividing  the  percentage  by  the  rate  : 
and  25  bu.  divided  by  .05  =  500  bu. 

Explanation.— If  5%  of 
any  number  of  bu.  =  25  bu., 
100%  must  equal  if^  of  25 
bu.,  or  20  times  25  bu.  = 
500  bu.  Ans. 

Observe  that  we  obtain 
this  result,  1st,  By  multi- 
plying the  25  bu.  by  the 
ratio  of  100%  to  5%  ;  2d,  By  dividing  the  25  by  the  reciprocal  of  the 
same  ratio,  which  is  the  rate  %  expressed  as  a  decimal,  or  simply 
the  rate.    Hence,  the 

302.  Rule. — I.  Divide  the  given  percentage  hy  the  rate. 

Or, 

II.  Multiply  the  percentage  hy  the  ratio  of  100%  to  the 
given  rate  %. 


2d  solution. 

^=500.     Or, 

5% 

5-^100  =  1  =  ^0*^-     Or- 
5%  :  100%  : :  25  bu.  :  (500)  bu 


500  PEECENTAGE. 


PROBLEMS 


1.  Of  what  is  30  da.,  lb%  ?  Ans.  200  da. 

2.  Of  what  is  1  ch.,  1^%  ?  Ans,  1  mi. 

3.  Of  what  is  1  pt.,  12^^?  Ans.  1  gal. 

4.  Of  what  is  $8.50,  16f  ^  ?  Ans.  $51. 

5.  Of  what  is  $1000,  Q^%  ?  Ans.  $1600. 

6.  Of  what  is  ^^  |^  ?  Ans.  $1.33^. 

7.  Of  what  is  l^^,  i%  ?  ^^s-  ^18}. 

8.  Of  what  is  £f ,  337^^  ?  Ans.  2|4s. 

9.  A  farmer  raised  40  bu.  oats,  which  was  25^  of  his  crop 
of  wheat.     How  much  wheat  did  he  raise  ?     Ans.  160  bu. 

10.  I  sold  for  $325  J,  1^%  of  my  interest  in  a  flouring  mill. 
What  is  the  value  of  my  remaining  interest  ? 

Ans.  $4014.50. 

11.  From  1863  to  1869,  the  average  length  of  the  school 
term  in  Pennsylvania  increased  .608  mo.,  which  was  equal  to 
11.2^  of  the  average  length  of  term  in  1863.  What  was 
the  length  of  that  term  ?  Ans   5.43  mo. 

12.  Having  sold  240  sheep  at  one  time,  and  210  at  an- 
other, and  120  at  another,  I  find  that  I  have  only  b%  of  the 
original  number  left.     How  many  sheep  had  I  at  first  ? 

Ans.  600  sheep. 

13.  A  man  who  owned  37^^  of  a  salt-works,  sold  33-^^  of 
his  interest  for  $756. 33J.  At  what  price  was  the  entire  works 
valued?  Ans.  $6050.66. 

14.  A  sold  f  of  62i^  of  his  U.  S.  Bonds,  for  $5000.  What 
is  the  value  of  those  he  retained  ?  Ans.  $5000. 


PERCENTAGE, 


301 


CASE      IV. 

303.   Sum  or  Difference  and  Rate  Per  Cent 
given  to  find  the  Base, 


©i^al  ^E:^Grx^iXGX-  '"^ 


Example  1. — What  number  is  that  which  if  increased 

by  20^  of  itself,  equals  48  ? 

Solution. — 20%  of  a  number  =  \  o{  the  number,  whicli  added  to 
f  or  tbe  number  =  f  of  the  number,  or  48. 

Since  48  is  f  of  the  number,  \  of  the  number  is  \  of  48,  or  8  ;  and 
f ,  or  the  number,  is  5  times  8,  or  40.  Therefore,  40  is  that  number  to 
which  if  20  %  be  added  the  sum  is  48. 

Example  2. — What  number  is  that  which  diminished  by 
12i^of  itself,  equals  56  ? 

Solution. — 12^%  of  a  number  =  i  of  the  number,  which  taken 
from  f ,  or  the  number  =  |  of  the  number,  or  56. 

Since  56  is  |  of  the  number,  |  of  the  number  is  ^  of  56,  or  8  ;  and 
I,  or  the  nimiber,  is  8  times  8,  or  64.  Therefore,  64  is  that  number 
from  which  if  12|%  be  taken  the  difference  is  56. 

PJtOBT^EMS. 


What  number  increased  by 

1.  20^  of  itself  =    24  ? 

2.  Vl\%  of  itself  =    72  ? 

3.  334^  of  itself  =  120? 

4.  16f  ^  of  itself  =    35  ? 

5.  150^  of  itself  =    25  ? 


What  number  decreased  by 

6.  25^  of  itself  =  100  ? 

7.  66|%  of  itself  =    9^  ? 

8.  4:%  of  itself  =    24  ? 

9.  ^%  of  itself  =  132  ? 
10.  374^  of  itself  =  200  ? 


11.  My  crop  of  beets  this  year  was  42  bu.,  which  was  b% 
more  than  last  year.     What  was  it  last  year  ? 

12.  I  harvested  this  year  450  bu.  of  wheat,  which  is  12^^ 
more  than  my  bins  hold.     What  is  their  capacity  ? 

13.  A  man  lost  20^  of  his  sheep,  and  had  120  left.     How 
many  sheep  had  he  at  first  ? 


302  PERCENTAGE. 

14.  Charles  attended  school  5  days  in  one  week,  which 
was  66f^  better  than  his  brother  James..  How  many  days 
did  James  attend  ? 

15.  A  man  bought  stock  to  the  amount  of  14000,  which 
lacked  20^  of  being  all  the  money  he  had.  How  much 
money  had  he  ? 

16.  A  farmer  purchased  a  piece  of  land,  containing  45  A., 
which  was  just  16f  ^  less  than  his  farm.  How  many  acres 
in  the  farm  ?  - 

17.  A  merchant  sold  his  bookkeeper  an  interest  in  hia 
business  for  $4000,  and  then  found  that  his  own  interest  was 
250^  more  than  that  of  his  bookkeeper.  What  was  the 
yalue  of  his  interest  ?    Of  both  their  interests  ? 


t  Wi^itten  ^x^erci^esTt 


Example  1.  —  What  number  increased  by  29%  of  itsell 
equals  1677? 

1st  solution.  Explanation.— Since  the  number 

1  -|-     .29  =    1.29  ^s  increased  by  29%  or  .29  of  itself, 

1fi77  _JL_  1  9Q  1^00  "^^^^  ^^  129%  or  1.29  times  tlie  num- 

ber:  and  1677  divided  by  1.29  gives 
the  quotient  1300,  or  the  number  required,  or  base. 

Explanation. — Represent- 
2d  solution.  .^^  ^^^  g^g^  ^y  ^QQ  ^^  ^  ^g  ^^^^ 

100^  4-  29^  =  129%  ;  the  entire    number,   1677   = 

1677  -7-  129  =  13  100%  +  29%  =  129%. 

13    X  100  =  1300,  Ans,  Now,  if  129%  =  1677,  1% 

must  equal  j^j  of  1677,  or  13, 
and  100%  =  100  times  13,  or 


Or, 


^^^^  X  100  =  ^  =  1300,  Ans.  1300 


129  1.29 

Or,  to  change,  by  cancellation,  the 

129%  :  100%  :  :  1677  :  (1300  )  129  to  the  form  1.29,  and  divide 


But,  it  is  more  convenient 
change,  by  ca 
9  to  the  form  1 
by  this  number. 


PERCE  ]S^T  AGE.  303 

Example  2.  —  What  number  diminished  by  31^  of  itself 

equals  1587? 

Explanation.— Since  the  number  is 

1st  solution.  diminished  by  31  f^  or  .31  of  itself,  1587 

1  —  .31  =      .69  is  69%  or  .69  of  the  number;  and  1587 

X537  _:_  .69  =  2300  divided  by  .69  is  2300,  or  the  number 

required,  or  base. 

Explanation.— 
2d  solution.  -  The    Base,    100 /e, 

100^  -  31^  =  6d%;  -^^^'  =  ^^^^  = 

1587  -^  69  =  23  ;  23  X  100  =  2300,  Ans.  ^^^'^^^  ^^  ~  ^fj^ 
Or,  100%  =  100  times 

1587  1587  23,   or    2300.      Or, 

-gg-  X  100  =^^^  =  i^300,  Ans.  ^^  cancellation,  we 

change  69  to  .69  and 

'  divide,  obtaining  in 

69^  :  100^  : :  1587  :  ( 2300 ).  both  cases  the  same 

result,  2300. 

304.  Rule.— I.  Divide  the  given  number  by  1  increased 
or  diminished  by  the  given  rate.     Or, 

II.   Take  such  a  part  of  the  given  number  as  100%  is  of 
the  sum  or  difference  of  100%  and  the  given  rate  %). 

PR  O  B  L  EMS. 

What  number 

1.  Increased  by  13^  of  itself  =  1582  ?  Ans.  1400. 

2.  Increased  by  17^^  of  itself  =  2350  ?  Ans.  2000. 

3.  Increased  by  22^  of  itself  =  366  ?  Ans.  300. 

4.  Increased  by  75^  of  itself  =  787.5  ?  Ans.  450. 

5.  Increased  by  125^  of  itself  =  833^  ?  Ans.  370^^. 

6.  Increased  by  325^  of  itself  =  2210  ?  Ans.  520. 

7.  Increased  by  22^  of  itself  is  1098? 


304  PERCENTAGE. 

What  number 

8.  Diminished  by  16|^  of  itself  =  700  ?         Ans.  840. 

9.  Diminished  by  75  %  of  itself  =  125  ?        Ans.  500. 

10.  Diminished  by  13^%  of  itself  =  260  ?  A7is.  300. 

11.  Diminished  by  62^%  of  itself  =  131.25  ?  Ans.  350. 

12.  Diminished  by  87^%  of  itself  =  28^  ?  Ans.  225. 

13.  Diminished  by  93i^  of  itself  r=  12|-  ?  Ans,  200. 

14.  A  drover  bought  360  head  of  cattle,  which  was  25^ 
more  than  he  already  had.  How  many  had  he  before  the 
purchase  ?  Ans.  288  head. 

15.  The  average  school  term  in  Pa.,  in  1869,  was  6.04  mo., 
which  was  11.2^  more  than  in  1863.  What  was  the  average 
term  in  1863  ?  Ans.  5.43  mo. 

16.  The  average  Pa.  school  term  in  1863,  was  5.43  mo., 
which  was  10.05^  less  than  in  1869.  What  was  the  length 
in  1869  ?  Ans.  6.04  mo. 

17.  The  distance  by  the  P.  R.  R.  from  Pittsburgh  to 
Harrisburg  is  246  miles,  which  is  125ff^  more  than  the 
distance  by  the  same  road  from  Harrisburg  to  Philadelphia. 
What  is  the  latter  distance  ?  Ans,  109.2  mi. 

18.  W.  R.  Ford  bought  a  farm  for  a  certain  sum,  ex- 
pended for  stock  11^  of  the  price  of  the  farm,  and  found 
that  the  whole  cost  was  17215.  What  was  the  cost  of  the 
farm  alone  ?  Aris.  $6500. 

19.  H.  J.  Gourley  raised  800  bu.  wheat,  which  was  25^ 
more  than  ^  of  what  Logan  raised.  How  many  bushels  did 
Logan  raise?  A7is.  1280  bu. 

20.  A  man  dying  bequeathed  10^  of  his  property  to  his 
son,  20;^  of  the  remainder  to  his  daughter,  33^;^^  of  this 
remainder  to  his  wife,  and  the  remainder,  or  $48000,  to 
charitable  purposes.     Required  the  value  of  the  estate. 

Ans,  $100000. 


PERCENTAGE.  305 

305.  Formulas, 

The  entire  subject  of  Percentage  may  be  summed  up  in 
the  following  formulas: 


Let  r  represent  the  rate. 
"  S       "  "  amount 

"   D       "  «  difference. 


Let  B  represent  the  base. 
"  F        "         "  percentage. 
"  R        «         "  rate  %. 


Then  the  Kule  in  Case  I.  becomes 
=  B  ^r,  or  p  =  B  x-^^. 

Case  II.  Case  IIL 


r  =:  -j^  and  ^  = 


^  X  i^^^.         B  =-  and  5=^  X  P. 
B  ^  r  R 


Case  IV. 

J,  S  D 

B  =  -— - — ,    or 


1  4-  r'  1  —  r 

A  thorougli  knowledge  of  the  use  of  these  formulas  is  about  all  the 
pupil  needs  to  carry  him  through  the  Applications  of  Percent- 
age ;  for  by  a  simple  substitution  of  the  proper  terms  for  B,  P,  R, 
&c. ,  they  can  be  made  applicable  to  all  cases  except  Interest,  in  which 
the  element  time  {t.)  is  introduced. 

306.  Applications  of  Percentage. 

Percentage  is,  because  of  its  convenience,  extensively  used 
in  business  calculations.  Its  principal  commercial  applica- 
tions are  in  Profit  and  Loss,  Insurance,  Com- 
mission, Interest,  Discount,  Stocks,  Exchange^ 
Taxes,  Partnership  and  Bankruptcy. 

20 


SECTION    III 


»MciFiT'  Mmm  iMmm 


'^^^•>0 


307.  Profit  and  Loss  are  terms  used  to  express  the 
gains  or  losses  in  business  transactions.  Gains  and  losses 
are  computed  at  a  certain  rate  per  cent  of  the  cost  or  the 
sum  invested,  and  the  operations  involve  the  2jri7icijples 
already  explained  in  Percentage. 

308.  The  Base  is  the  cost  or  the  sum  invested. 

309.  The  Kate  %  is  the  rate  per  cent  of  profit  or  loss. 

310.  The  IPercentage  is  the  profit  or  loss. 

311  {a).  The  Amount  is  the  sm7?2  of  cos^  and  j^rq/?^,  or 
the  selling  price. 

311  (5).  The  Difference  is  the  co5^  Zess  the  Zo55,  or  the 
selling  price. 

CASE    I. 

312.  To  find  the  sum  gained,  or  lost,  when 
the  cost  and  rate  %  are  given. 


■■♦•- 


—{i 


■\-  €)i-alE^oicisG< 


Example  1.— A  wagon  that  cost  $150  was  sold  at  a  gain 
of  20,^.     What  was  the  gain  ? 

Solution.— Since  the  gain  was  20%,  it  was  ^,  or  \  of  $150, 

or  $30. 


316 


PROFIT     A  KD     LOSS. 


307 


Example  2. — A  horse  that  cost  1300  was  sold  at  a  loss  of 

(^f-     What  was  the  loss  ? 

Solution.— Since  the  loss  was  27%,  it  was  ^^^  of  $300,  or  $81. 


l*JtOBL]£MS. 


How  much  is  gained  or  lost  when  the  cost 
1.  Is  $50,  gain  20^  ?  6.  Is  $500,  loss  12^? 


2.  Is  $67,  loss  10^  ? 

3.  Is  $800,  gain  30^  ? 

4.  Is  $1000,  loss  15^  ? 

5.  Is  $30,  gain 


7.  Is  $800,  gain  50^? 

8.  Is  $600,  loss  40%  ? 

9.  Is  $765,  gain  100%  ? 
10.  Is  $250,  gain  75%  ? 


-.-♦-- 


t  ^i^itten  ^X^^er  c  i;^e^  i: 


SOLUTION. 

X  ^  =  $37.50,  gain. 

2 

Or, 
$250  X   .15  =  $37.50,  gain. 


Example. — What  is  the  gain,  when  goods  are  bought  for 
$250  and  sold  at  a  gain  of  15%  ? 

Explanation.— The  gain  is 
15%  (R,  in  Percentage),  or  ^J^ 
of  the  cost  (B),  which  is  found 
either  by  multiplying  the  cost 
by  the  number  denoting  the 
rate  % ,  and  dividing  this  pro- 
dact  by  100 ;  or  by  multiplying 

the  cost  by  the  rate  (r),  .15.     In  either  case  we  obtain  the  same  result, 

$37.50. 

313.   EuLES. — Take  such  part  of  the  cost  as  the  rate  % 
is  of  100%.     (A  X  ^  =  P.)     Or, 
Multiply  the  cost  hy  the  rate.     {B  x  r  ^  P.) 

I*  ROBLEMS. 

1.  Sold  goods  which  cost  me  $1250,  at  a  gain  of  12|^%. 
How  much  did  I  gain?  Ans.  $156.25. 


308  PERCENTAGE. 

2.  How  much  do  I  lose  by  gelling  a  horse,  which  cost  me 
$542,  at  a  loss  of  18^  ?  Atis.  $97.56. 

3.  I  wish  to  sell  my  goods  at  a  profit  of  20%;  how  shall 
I  mark  calicoes  that  cost  10)^  ?  Delaines  that  cost  40^  ? 
Towehng  that  cost  12^^  ?  Silks  that  cost  $3  ?  Broadcloths 
that  cost  $4.50  ?     Latins  that  cost  $5  ?      Ans.  to  last,  $6. 

4.  I  am  closing  out  my  stock,  and  willing  to  lose  12^^  on 
cost  of  my  goods.  How  should  I  mark  hose  that  cost  3'Z^  ? 
Collars  that  cost  8^  ?     Cloaks  that  cost  $32  ? 

Aiis.  to  last,  128. 

CASE    II. 

314.  To  find  the  rate  %  of  gain  or  loss,  when 
the  cost  and  gain,  or  loss,  are  given. 


4^  ©ral^:<^GFciXe, 


Example. — Bought  a  shawl  for  $16,  and  sold  it  at -an 
advance  of  $4.    What  %  did  I  gain  ? 

Solution. — My  gain  is  $4  on  $16,  or  y\  =  ^  of  cost. 
Since  cost  is  100  % ,  my  gain  is  \  of  100  % ,  or  25  % . 

mo  B  LEMS. 

1.  Bought  calico  for  8^  per  yard,  and  sold  it  so  as  to  gain 
2^  on  each  yard.     What  %  did  I  gain  ? 

2.  Bought  shoes  at  $5,  and  sold  them  at  $6.     What  was 
the  gain  %  ? 

3.  Bought  oranges  at  3f  apiece,  and  sold  them  at  bf . 
What  is  the  rate  %  gain  ? 

4.  Bought  apples  at  3  for  bf ,  and  sold  them  at  4  for  bf . 
What  %  was  the  loss  ? 

5.  Bought  a  book  for  $2,  and  sold  it  for  $1.50.    What 
was  the  loss  %  ? 


PROFIT     AND     LOSS.  309 


-•>. 


^IWritten  ^xTerei^er-i 


Example.— Bought  a  buggy  for  1360,  and  sold  it  for  $450. 
What  was  the  rate  %  of  gain  ? 

SOLUTION. 

$450,  selHng  price.  Explanation.— By  subtraction  we  find 

360,  cost.  the  gain  is  $90.     This  is  gained  on  $360 

7               T~  (B).     The  rate  %  (R),  therefore,  must  be 

$  90,  gain.  ^5H,^  Qj.  1^  ^f  -^QQ  a/^  =  25  % .    Or,  by  dividing 

-//jp  of  100^  =  25^.         the  gain,  $90  (P),  by  the  cost,  $360  (B),  we 
Or,  obtain  the  same  result,  .25  =  25%. 

$90-^1360=1.25  =  25^. 

315.  Rules. — Talce  such  a  part  of  100%  as  the  gain  or 
loss  is  of  the  cost.    (b  =  -j^x  100%.\     Or, 
Divide  gain  or  loss  by  cost.    (P  -r-  B  =  r.) 

1.  Bought  a  house  for  $1750,  and  sold  it  for  $500.  What 
^  is  the  loss  ?  Ans.  71-|^. 

2.  Bought  a  farm  for  $7650,  and  sold  it  for  $9780.  Re- 
quired the  gain  %-  A7is.  27f|^. 

3.  Bought  a  horse  for  $160,  and  sold  him  for  $25.  What 
^  did  I  lose?  Ans.  84|^. 

4.  Bought  delaines  for  20^  a  yard,  and  sold  them  at  37J^  a 
yard.     Required  the  rate  %  gain.  Ans.  87^-^. 

5.  Bought  100  doz.  eggs  at  15)^  a  doz. ;  but  20  doz.  were 
broken  in  transportation.  Sold  the  remainder  at  Slf  a  doz. 
What  did  I  gain  on  the  investment  ?  Ans.  65^%. 

6.  I  buy  a  horse  for  $100,  and  after  keeping  him  at  a 
livery  stable  for  10  weeks  at  an  expense  of  $5  a  week,  I  sell 
him  for  $120.     What  is  my  rate  %  of  loss  ?  Ans.  20^. 


310 


PEKCENTAGE. 


CASE    III. 

316.  To  find  the  cosf^  when  the  gain,  or  loss, 
and  the  rate  %  are  given. 


Qved  f^Xoi^cj^h 


K 


Example. — Sold  my  horse  at  an  advance  of  125,  which 
was  20%  of  his  cost.     What  was  his  cost  ? 

Solution. — Since  20  %,  or  y%"|j,  or  |,  of  the  cost  is  $25,  f ,  or  the  cost, 
Is  5  times  $25,  or  $125. 

PROBLEMS. 

1.  If  $20  is  20%  of  the  cost  of  16  bbl.  of  flour,  what  is  the 
entire  cost  ? 

2.  If  I  lose  $16  on  the  price  of  an  ox,  my  loss  amounts  to 
12|^%  of  his  value.     What  is  his  value  ? 

3.  If  $10  is  5%  of  the  cost  of  my  fare  to  San  Francisco, 
what  is  the  whole  cost  ? 

4.  A  lost  $450,  which  was  90%  of  all  he  had.  How  much 
did  he  have  ? 

5.  A  grocer  sold  some  bacon  for  $250,  which  was  equal 
to  12^%  of  the  value  of  all  he  had.  What  was  the  entire 
value  ? 


-•>•- 


t^  1  itten  ^?&i^ci;^es 


t. 


r- 


Example. — I  sold  my  farm  at  a  gain  of  $500,  which  was 
equal  to  20%  of  its  value.     What  was  its  value  ? 


PROFIT     AlfD     LOSS.  311 

Explanation. — The    whole  value 

SOLUTION.  of  the  farm  is  100^  of  it.     But  100% 

<i&KnA  _  ^9nnn         is  5  times  20%  ;  and  as  $500  is  20;^ 

^  $2500,  is  its  value. 

'  Or,  since  $1  gains  $.20,  it  vrill  re- 

$500  ~  .20  =  12500.  quire  as  many  dollars  to  gain  $500  as 

.20  is  contained  times  in  500. 

Sm,  Rules. — Take  such  a  part  of  the  gain  or  loss  as 
100%  is  of  the  given  rate  %.     (-^    x  P  =  B.\     Or, 
Divide  the  gain  or  loss  hy  the  rate.     {P  -^  r  =  B). 

r  ROBLEMSc 

1.  What  was  the  cost  of  goods,  when  a  loss  of  110  was  6% 
of  the  cost?  •  ^W5.  $200. 

2.  What  was  the  cost  of  goods,  when  a  gain  of  $315  was 
2J^  of  cost  ?  Ans.  $14000. 

3.  I  sold  my  house  and  lot  for  $7500,  which  was  150^  of 
its  cost.     What  was  its  cost  ?  Ans.  $5000. 

4.  Three  hundred  thousand  drafted  men  were  equal  to 
62J-^  of  the  number  called  for  by  the  government  in  1  year. 
What  was  the  whole  number  called  ?     Ans.  480000  men. 

5.  By  selUng  my  horse  and  chaise  at  lb%  of  their  cost,  I 
received  for  them  $375.    What  did  they  cost?  Ans.  $500. 

CASE    IV. 

318.  To  find  the  cost,  when  the  selling  price 
and  rate  %  of  gain,  m^  loss,  are  given. 


Oi*al  ^x:orei^eX 


Example  1. — By  selling  hats  at  $6  apiece,  I  gain 
What  do  the  hats  cost  me  ? 


312 


PERCENTAGE. 


Solution. — Since  the  gain  is  20%  ,or  ^,  of  the  cost,  the  selling  price 
is  the  cost  plus  i  of  cost,  or  f  of  the  cost.  Since  $6  is  f  of  the  cost, 
I  of  $6,  or  |5,  is  the  cost. 

Example  2. — A  knife  was  sold  for  87^^^,  which  was  12^% 
less  than  it  cost.     Kequired  the  cost. 

Solution. — Since  the  loss  is  12\%,  or  \  of  the  cost,  the  selling 
price  is  the  cost  less  |  of  the  cost,  or  |  of  the  cost.  Since  87^  cents, 
or  $|,  is  I  of  the  cost,  f  of  $|,  or  $1,  is  the  cost. 

JPB,  OBLEMS. 

1.  Sold  a  pencil  for  IQf,  which  was  25^  more  than  its 
cost    What  was  the  cost  ? 

2.  Sold  calico  at  Sf  a  yard,  which  was  at  a  loss  of  20^. 
"What  was  its  cost  ? 

3.  Bought  hay  at  $12  a  ton,  which  was  20^  less  than  the 
market  price.     What  was  the  market  price  ? 

4.  Bought  potatoes  from  store  at  40^  per  bu.,  which  was 
at  an  advance  of  33^^  on  farmers'  prices.  What  were  farm- 
ers' prices  ? 

5.  I  pay  40^  a  lb.  "the  year  round"  for  butter,  which  is 
60^  more  than  my  neighbors  pay  in  summer,  and  33^^ 
less  than  they  pay  in  winter.  What  is  the  average  price 
paid  by  my  neighbors  ? 


-.-♦».- 


:i^s>Yi^ittei\^:xrerei^esr^ 


Example. — By  selling  my  farm  for  $4000, 1  gained  20^. 
What  did  my  farm  cost  ? 

SOLUTION. 

100^  +  20%  =  120%  =  $4000. 
$4000  -T-  120  =  $33^  =  1%. 
1%  X  100  =  $33^  X  100  =  $33334. 
Or, 


$4000  -J-  (1  4-  .20  =  1.20) 


$33334. 


Explanation. — We 
first  add  the  gain, 
20%,  to  the  cost, 
100%,  to  find  what 
fo  of  the  cost  =  the 
selling  price,  and  ob- 
tain   120%     of    cost. 


PROFIT     AND     LOSS.  313 

We  then  divide  the  $4000  by  120  to  find  l^o  of  the  cost,  or  $33f    This 
we  multiply  by  J  00  to  find  100  5^  of  the  cost,  or  the  whole  cost=|3333i-. 
Or,  as  we  receive  $1.20  for  each  $1  invested,  we  must  have  in- 
vested as  many  dollars  as  $1.20  is  contained  times  in  $4000. 

319.  KuLES. — Take  stick  part  of  the  selling  price  as  100% 
is  of  100%  +  the  rate  %  of  gain,  or  —  the  rate  %  of  loss. 

Divide  the  selling  price  hy  1  increased  hy  the  rate  of 
gain,  or  diminished  hy  the  rate  of  loss.     I  =  B,  or 


47=-) 


1  —  r 

FBOBZEMS. 

What  is  the  cost  of  goods  which  selling 

1.  At  a  loss  of  20^  bring  $30  ?  Afis.  $37.50. 

2.  At  a  gain  of  25%  bring  $75  ?  Ans,  $60. 

3.  At  a  gain  of  12|^  bring  $.72  ?  Ans.  $.64. 

4.  At  a  loss  of  8 J;;^  bring  $132.11  ?  Ans.  $144.12. 

5.  Sold  a  watch  at  33^^  gain,  and  with  the  money 
bought  another,  which  I  sold  for  $120,  and  lost  20^.  Did 
I  gain  or  lose,  and  how  much  ?  Ans.  $7.50  gain. 

6.  Sold  a  wagon  at  $250,  and  thereby  gained  11^.  What 
did  the  wagon  cost  ?  Ans.  $225.23, 

MEVIEW    PROBLEMS, 

1.  How  must  goods  be  marked  which  cost  24/,  so  as  to 
take  off  16|^,  and  still  make  25^  ?  A7is.  36f . 

Note. — First  find  selling  price  at  25  %  gain  (Case  I) ;  then  regard 
the  selling  price  as  cost  and  find  asking  price  (Case  IV), 

2.  Find  the  asking  price  when  cost  is  $1.20,  so  as  to  fall 
6J^,  and  still  make  25^.  Ans.  ^l.QO. 


314  PERCENTAGE. 

3.  Cost  $1.50 ;  from  what  price  can  I  fall  10^,  and  still 
make  20^  ?  Ans.  12. 

4.  Cost  9^ ;  from  what  price  can  I  fall  10^,  and  still  make 
50%?  Ans.lbt 

5.  Cost  20)^ ;  from  what  price  can  I  fall  25%,  and  still  make 
a  profit  of  20%  ?  Ans,  32^. 

6.  If  goods  selling  at  $2.40,  realize  33-J%  gain,  what  would 
be  the  rate  %  of  gain  at  $2.10  ?  Ans.  16|%. 

Note. — First  find  cost  (Case  IV.) ;  then  rate  %  of  gain  (Case  II.) 

7.  If  sales  at  $1.80  make  10%  loss,  what  would  be  the  rate 
%  of  gain  at  $2.20  ?  Ans.  10%. 

8.  If  property  at  $800  makes  20%  loss,  what  would  be  the 
rate  %  loss  at  $750  ?  Ans.  25%. 

9.  If  a  bank  share  at  $62.50  is  25%  gain,  what  would  be 
the  rate  %  loss  at  $48  ?  Ans  4%. 

10.  If  land  at  $56  an  acre  is  12%  gain,  what  would  be  the 
rate  %  gain  at  $60  an  acre  ?  Ans.  20%. 

11.  If  nails  at  6<^  a  lb.  make  4%  loss,  what  would  be  the 
rate  %  gain  at  Hf  a  lb.  Ans.  12%. 

12.  If  a  merchant  realizes  20%  gain  on  tea  at  75^  a  lb., 
what  would  he  do  at  56^^^  a  lb.  ?  Ans.  Lose  10%. 

13.  Selling  price  15f ;  gain  25% ;  find  rate  %  of  gain  at 
16;^.  Ans.  33-Jr%. 

14.  Selling  price  16 f;  loss  25%;  find  the  rate  %  of  gain 
at  25^.  Ans.  25%. 

15.  A  speculator  sold  2  houses  at  $6000  each  ;  on  one  he 
gained  33  J%  of  his  investment,  and  on  the  other  he  lost 
33-J-%  of  his  investment.  Did  he  gain  or  lose,  and  how 
much?  Ans.  $1500  loss. 

16.  A  man  sold  a  house,  which  was  presented  to  him,  for 
$5000.    What  was  his  rate  %  of  profit?  Ans.  ? 


^^^^:^ 


SECTION    IV 

4 


IMSTUMJtMC'E 


^fc^ 


320.  Insurance  is  indemnity  against  pecuniary  loss, 
pledged  by  one  party  to  another. 

321.  The  Premium  is  the  sum  paid  for  insurance, 
and  is  a  certain  per  cent,  of  the  sum  insured. 

322.  The  Policy  is  the  contract  between  the  parties. 

323.  Fire  Insurance  is  indemnity  for  loss  by  fire. 

324.  Marine  Insurance  is  indemnity  for  loss  by 
navigation. 

325.  Health  Insurance  secures  an  allowance  of 
money  during  the  sickness  or  disability  of  the  party  in- 
sured. 

326.  Life  Insurance  secures  a  specified  sum  of 
money  to  a  designated  person,  or  to  designated  persons,  or 
to  their  heirs,  on  the  death  of  the  party  iiisured,  or  on  his 
arriving  at  a  certain  age. 

327.  Underwriters  or  Insurers  are  the  parties 
that  insure  against  loss. 

328.  Fire  and  Marine  Insurance, 

Fire  and  Marine  Insurance  are  so  completely  covered  by  Cases  I, 
11,  and  III  of  Percentage,  that  it  is  thought  unnecessary  to  do  more 
than  say  that  the 

Value  Insured  corresponds  to  the  base  .    .    .    -B, 
Premium  "  "    "    percentage.    P. 

Bate  Per  Cent,  **  "     '*    rate  %  .    .    B, 

815 


316  PEBCENTAGE 


CASE    I. 

329.  To  find  the  I^remiuin^  when  the  Value 
and  Kate  %  are  given. 

l*lt  O  BLEM  S. 

What  is  the  Premium  on 

1.  A  house  insured  for  $2000,  at  1^%?  A^is.  $30. 

2.  A  store  iusured  for  $3500,  at  2^?  A71S,  $70. 

3.  A  barn  insured  for  $900,  at  1%  ?  Ans.  $6.75. 

4.  Furniture  insured  for  $1250,  at  |^  ?  Ans.  $7.81. 
.5.  A  church  insured  for  $16000,  at  H%  ?  Ans.  $300. 

6.  A  factory  insured  for  ^64000,  at  2^^?  Ans.  $1440. 

7.  A  ship  insured  for  $15000,  at  3^?  Ans.  $450. 

8.  A  steamboat  insured  for  $85000,  at  2-J;^?  Ans.  $2125. 

9.  Freight,  for  $3780,  at  11%  ?  Ans.  $66.15. 

10.  A  mill,  for  $12500,  at  ^%  ?  Ans.  $312.50. 

11.  Clothing,  for  $875,  at  {%  ?  Ans.  $7.66. 

CASE    II. 

330.  To  find  the  Rate   %,  when  the    Value 
and  Premium  are  given. 

PMO  B  TjEMS. 

What  is  the  Rate  %,  when  the 

1.  Value  insured  is  $500,  premium  $6  ?  Ans.  1.%%. 

2.  Value  insured  is  $1000,       "        $8?  Ans.  .H%. 

3.  Value  insured  is  $800,         "        $3  ?  Ans.  %%. 

4.  Value  insured  is  $3600,        "        $56  ?  Ans.  1^%. 

5.  Value  insured  is  $5000,        "        $125  ?  Ans.  2^%. 

6.  Value  insured  is  $8000,       "        $280  ?  Ans.  3^%. 

7.  Value  insured  is  $1800,       "        $24  ?  Ans.  1^%. 


INSURANCE.  317 

CASE    III. 
331.   To  find  the  Value  Insured^  when  the 
Rate  %  and  J^remium  are  given. 

jPiJ  O  BLEMS. 

What  is  the  value  insured,  when  the 

1.  Premium  is  $6  and  1.2  the  rate  %  ?  Arts.  $500. 

2.  Premium  is  18  and  .8  the  rate  %  ?  Ans.  $1000. 

3.  Premium  is  $3  and  |  the  rate  %  ?  Ans,  $800. 

4.  Premium  is  $24  and  1|  the  rate  ^  ?  Ans.  $1800. 

5.  Premium  is  $56  and  1|  the  rate  %  ?  Ans.  $3600. 

6.  Premium  is  $125  and  2^  the  rate  ^  ?  Ans.  $5000. 

7.  Premium  is  $280  and  ^  the  rate  %  ?  Ans.  $8000. 

332.  Life  Insurance. 

Life  Insurance  is  of  different  kinds. 
The  rates  paid  for  life  insurance  depend  upon  the  age  of 
the  party  insured,  and  the  kind  of  Policy  taken  out. 

For  example, — it  costs  more  per  year  to  insure  a  man  50  yr.  of  age, 
than  one  40  yr.  of  age. 

It  also  costs  more  per  yr.,  to  insure  a  man  of  any  number  of  years 
of  age,  when  the  payments  are  to  be  made  in  a  stated  number  of  years, 
than  when  they  are  to  be  made  every  year  he  lives. 

It  also  requires  a  higher  rate  for  any  given  man  when  the  insur- 
ance is  to  be  paid  at  the  end  of  some  stipulated  period,  if  the  man  be 
living,  or  at  his  death  should  it  previously  occur. 

PR O  B  LEMS  . 

1.  A  man  at  the  age  of  53,  paid  at  the  rate  of  5.38^  per 
year,  for  an  insurance  on  his  hfe.  What  was  the  annual 
premium  on  ^15000  ?  Ans.  $269. 

2.  After  having  paid  an  annual  premium  of  2.6^  on 
$10000  for  5  yr.,  the  insured  died.  How  much  more  than 
was  paid  was  received  by  the  heirs  ?  Ans.  $8700. 

3.  A  man  paid  for  40  years  a  premium  of  2.53^  on  $7500. 
How  much  did  he  pay  in  all  ?  Ans.  $7590. 


Qy 


6^ 


SECTION    V. 

K^Q>^^ 


CC>MMISSI©M  I' 


^ '^^^ . 

333.  Commission  is  the  pay  of  an  agent,  collector, 
broker,  &c.,  and  is  usually  estimated  on  the  amount  of  money 
he  receives  or  expends. 

334.  An  Af/ent  or  Factor  is  a  person  who  transacts 
business  for  another. 

335.  A  Commission  Merchant  is  a  merchant  who 
buys  and  sells  goods  for  others. 

336.  A  broker  is  an  agent  who  buys  and  sells  stocks, 
notes,  money,  &c. 

337.  SroUerage  is  the  commission  of  a  Broker. 

338.  A  Consignment  is  merchandise  sent  to  an 
agent  for  sale. 

339.  A  Consignor  is  a  sender  or  maker  of  a  con- 
signment. 

340.  A  Consignee  is  an  agent  who  receives  a  con- 
signment. 

Commission  embraces  the  Fo^ir  Cases  of  Percentage.  Hence,  the 
Bame  Rules  apply,  and  need  not  be  repeated. 

The  Proceeds  of  sales,  or  the  moneys  collected  or  invested 

correspond  to  the  tase S, 

The  Commission   corresponds  to  the  percentage  . .  P. 
The  Rate  %  "         «    "    rate  %   .  .  ,  .  iJ. 

The  Proceeds  +  Com.   "         '*    "    amount S* 

The  Proceeds  —  Com,   "         "    "   difference  . . .  D, 

318 


COMMISSION.  319 

CASE    I. 
341.  To  find  Corn/mission. 

PROBLEMS. 

What  is  the  commission  on  buying  or  selling 

1.  Flour  for  $7200,  at  3^  ?  Ans.  $216. 

2.  Oats  for  $2800,  at  ^%  ?  Ans.  $70. 

3.  Iron  for  $17500,  at  1^%  ?  A7is.  $262.50. 

4.  A  sells  B's  consignment  for  $1985,  at  3^  commission. 
How  much  does  A  remit  B  ?  Ans.  $1925.45. 

5.  0  buys  for  D  merchandise  worth  $13000,  at  2^^  com- 
mission.    How  much  should  D  send  C  ?        Ans.  $13325. 

6.  E's  broker  buys  for  him  $9000  worth  of  stocks,  at  1% 
brokerage.    What  does  E  pay  ?  Ans.  $9033. 75. 

7.  F's  broker  sells  for  him  a  note  for  $960,  at  ^%  broker- 
age.   How  much  does  F  get  ?  Ans.  $957.60. 

8.  A  consignee  sells  oil  for  $7250,  at  3J^  commission. 
What  does  the  consignor  receive  ?  Ans,  $6996.25. 

CASE    II. 
342.  To  find  Bate  %, 

P  It  O  BIjEM  s, 

1.  A  merchant  received  $153  for  selling  $5100  worth  of 
flonr.    What  %  commission  did  he  get  ?  Ans.  3%. 

2.  A  real  estate  broker  charged  $53.75  for  selling  a  fai-m 
for  $2150.    What  rate  %  was  his  commission  ?    Ans.  2^%. 

3.  A  broker  charged  me  $218.75  for  selHng  $25000  worth 
of  R  R.  stock.    What  %  was  the  brokerage  ?       Ans.  i%. 

4.  A  factor  charged  $230,625  for  selling  1025  bbl.  apples 
at  $5  a  bbl.,  and  $5.76  for  selling  256  bu.  potatoes  at  ^ 
per  bu.    What  was  the  rate  %  commission  ?       Ans.  A^%. 


320  PERCENTAGE. 

CASE    III. 

343.  To  find  Collections^  Purchases^  or  Pro- 
ceeds of  Sales 9  Commission  and  Kate  %  being 
given, 

PJt  O  BT.EMS, 

1.  Paid  an  agent  2^^  commission  on  the  sale  of  lands. 
His  bill  was  $524.50.    What  was  the  value  of  his  sale  ? 

Ans.  $20980. 

2.  A  real  estate  broker's  income  for  1875  was  $10563. 
His  commissions  were  2.1^.    What  were  his  sales  ? 

Ans.  $503000. 

3.  My  factor  at  Chicago  sent  me  a  bill  on  account  of  a 
consignment  of  grain  to  him,  for  $525,  which  included  his 
commission  at  3;^,  and  charges  for  freight,  drayage,  &c., 
$125.     What  was  the  value  of  the  consignment  ? 

Ans.  $13333^. 

Note. — Deduct  $125,  charges,  from  the  bill ;  the  remainder  is  the 
commission. 

4.  An  auctioneer  whose  commission  is  1%,  charged  $25 
for  selling  some  insurance  stock.  What  was  the  value  of 
the  stock  ?  Ans.  $2500. 

5.  A  collector  charged  me  $25  for  collecting,  at  b%.  How 
much  money  did  I  receive  ?  Ans.  $475. 

CASE    lY. 

344  (a).  To  find  Amount  of  Sale,  or  Collec 
tionSf  Mate  %,  and  Remittance  being  given; 
and, 

344  (b).  To  find  Investment,  Bate  %,  and  an 
Amount  covering  Investment  and  Commis- 
sion being  given. 

Note.— Apply  Rules  in  Case  IV  of  Percentage. 


COMMISSION.  321 

PR  O  B  T.EMS, 

1.  My  factor  remits  to  me  $3910,  the  proceeds  of  a  con- 
signment of  wheat,  after  deducting  his  commission  of  3^. 
For  what  did  he  sell  the  wheat  ?  Ans.  $3000. 

2.  1  sent  $5000  to  a  real  estate  broker  in  Chicago,  to  in- 
vest in  lands,  with  instructions  to  deduct  his  commission  at 
4:%.     What  sum  did  he  invest  ?  Ans.  $4807.69. 

3.  A  collector  in  Dubuque  remitted  to  me  $525,  as  due 
from  a  debt  he  had  collected,  after  deducting  h%  for  col- 
lecting.   What  was  the  debt  ?  Ans.  $552.63. 

4.  A  broker  having  intrusted  to  him  110000  to  invest  in 
stocks,  performed  the  duty,  charging  ^%  for  his  services. 
How  much  of  the  $10000  did  he  invest?     Ans.  $9913.26. 

5.  A  commission  merchant  sold  goods  to  the  amount  of 
$7560,  charging  3^  commission.  After  paying  $36.50 
charges,  he  invested  the  balance  in  coffee,  and  charged  2J^ 
for  the  investment.  If  he  paid  20^  per  lb.  for  the  coffee, 
how  many  lb.  did  he  purchase  ?  Ans.  35593.65  lb.  + 

6.  A  commission  merchant  sold  goods  to  amount  of  $6000, 
charging  1^%  for  his  commission  and  $16  for  freight.  How- 
much  did  he  remit  ?  Ans.  $5894. 

7.  A  broker  having  invested  $3600  in  stocks  for  me,  sent 
me  a  bill  for  commission  at  \^%  and  $3  for  telegraphing,  &c. 
I  sent  him  a  draft  for  $1000,  with  instructions  to  take  out 
his  former  bill,  and  invest  the  balance  in  bank  stock.  How 
much  did  he  invest,  after  deducting  a  commission  of  1%  on 
last  transaction  ?  Ans.  $933.66  +  . 

8.  I  sent  an  agent  $5922,  with  which  to  buy  stocks,  after 
deducting  a  commission  of  6%.  What  was  his  commission, 
and  how  much  did  he  invest  ? 

Ans.  Com.  $282;  invested,  $5640. 

21 


SECTION    VI 


■^lA^ 


(5^ 


IMT;E:ItE;ST' 


^i) 


345.  Interest  is  a  compensation  for  the  use  of  money. 

Interest  has  Five  Cases  ;  and  the  operations  do  not  differ  from  those 
in  Percentage,  except  in  respect  to  the  new  element,  Time. 

In  Interest,  however,  only  money  can  be  the  hose;  while  in  Percent- 
age the  base  may  be  anything. 

346.  The  JPrincipal  is  the  sum  of  money  for  the  use 
of  which  compensation  is  made. 

347.  The  Amount  is  the  sum  of  the  Principal  and  its 
Interest 

If  I  pay  $6  for  the  use  of  $100  for  1  year,  $6  is  the  interest;  $100 
the  princvpal ;  and  $100  +  $6,  the  amount. 

Interest  as  to  methods  of  reckoning,  is  of  two  kinds.  Simple  and 
Compound. 

348.  Simple  Interest  is  reckoned  on  the  Principal 
alone  at  a  certain  %  for  some  fixed  period,  generally  1  year. 

349.  Compound  Interest  is  reckoned  on  the  Prin- 
cipal for  the  first  period,  and  after  that  on  the  Principal  and 
Interest. 

Interest  as  to  rate  charged  is  of  two  kinds.  Legal  and  lUegal. 

350.  Legal  Interest  is  that  established  by  law. 

351.  Illegal  Interest,  called  Usury,  is  that  which 
exceeds  Legal  Interest,  and  is  not  allowed  by  law. 

The  taking  of  Usury  is  a  violation  of  law,  and  usually  has  attached 
to  it  some  penalty. 

325 


INTEREST. 


323 


351'.  The  Hate  IPer  Cent  of  Interest  is  the  number 
denoting  how  many  hundredths  of  the  Principal  equals  the 
Interest. 

The  laws  regulating  the  rate  per  cent,  and  the  forfeitures  and  pen- 
alties for  usury  are  different  in  different  States  and  Countries ;  in  the 
United  States  the  most  common  legal  rate  per  cent  is  6  % . 

The  following  table  will  afford  the  pupil  a  general  idea  of  the 


Legal  Mates  of  Interest, 


Legal  rate  in 

England 

France 

Louisiana 


Dakota 

Georgia  

Kentucky 

Michigan 

Minnesota 

New  Jersey 

New  York 

South  Carolina. 
Wisconsin 


Alabama , 
Texas... 


Arizona 

California... 
Colorado  — 

Idaho 

Kansas 

Montana  — 
Nebraska . . . 

Nevada  

Oregon 

Utah 

Washington. 

Wyoming  T. 


.10^ 


.12^ 


All  other  States . . . 
Dist.  of  Columbia. 
Debts  due  U.  S. . . . 

Canada 

Ireland 

Nova  Scotia 


Rates  allowed  in 

Arkansas  

California 

Missis!?ippi 

ISIassachusetts.. 

Nevada 

Rhode  Island.  ■ . 


. .  .Any  rate, hy 
agreement. 


Dakota.... 
Nebraska 


70 

15^ 


Minnesota 

Oregon 

Texas 


Florida 

Illinois..  ., 
Michigan. . 
Missouri . . 
Tennessee 
Wisconsin. 

Louisiana . 
Ohio 


10% 


Pennsylvania 
cial  law 
few  institutions 


lia,  by  spe-  ) 

in  cases  of  a  v  .  .  7% 

itutions ) 


Kansas,  do 


324  PEKCEKTAGE. 

CASE     I. 

353.  OChe  Principal^  Bate  %  and  Time  given^ 
to  find  the  Interest  and  Amount. 


0i*al  '^Xoi^ci^e'X 


Example  1. — At  6%  per  annum,  what  is  the  interest  of 

112  for  1  year  ? 

Solution.— If  the  interest  of  $1  for  1  yr.  is  6%,  it  is  6i^;  and  of 
$12,  it  is  13  times  as  much,  or  13  times  6^  =  73^2^.    Or, 

Since  6  %  is  yf  ^  =  /^^  of  the  principal,  the  interest  of  $13  for  1  year, 

at  6%,  is  /^  of  $13,  or  $.73. 

Example  2. — What  is  the  interest  of  $25  for  1  year,  at 
6%  ?    For  2  yr.  ?    For  3  yr.  ?    For  7  yr.  ? 

Solution. — At  5  %  the  interest  =  y^,  or  -^^  of  the  principal ;  ^  of 
$35  is  $1^,  or  $1.35.  For  3  yr.  it  is  3  times  $1.35,  or  $3.50  :  for  3  yr., 
3  times  $1.35,  or  $3.75 ;  &c. 

I^MOBLBMS. 

1.  What  is  the  interest  of  $50  for  3  yr.,  at  6^  ? 

2.  What  is  the  interest  of  $75  for  10  yr.,  at  b%  ?  At  6^  ? 
At  7^?    At  8^?    At  9^? 

3.  What  is  the  interest  of  $150  for  5  yr.,  at  b%  ?  At  6^  ? 
At  7^?    At  8^?    AtlO^^?    At  12^? 

Example  3. — What  is  the  interest  of  $100  for  1  yr.  3  mo., 
at  6^?    At  8^?    At  9^?    At  10^  ?    At  12^  ? 

Solution.— 1  yr.  3  mo.  =  1^»^,  or  1^,  or  f  yr.  The  interest  of  $100 
for  1  yr.  at  6%  is  $6 ;  therefore,  the  interest  for  f  yr.  at  6%  =  |  of 
$6,  or  $7^.  If  the  interest  at  6%  =  $7|,  at  8%  it  will  be  |,  or  |  of 
$7i  =  $10;  at  9%,  I,  or  f  of  $7^  =  $11^;  &c.,  &c. 

4.  What  is  the  interest  of  $200  for  2  yr.  9  mo.,  at  6^  ? 
At  8^?    At  9^?    At  10^?     At  12^? 

5.  What  is  the  interest  of  $500  for  1  yr.  3  mo.,  at  6^? 
At  7^?    At  10^?    At  15^? 


INTEREST. 


325 


Example  4.— What  is  the  amount  of  $20  for  2  yr.  6  mo., 

at  6^?     At  9^?    At  12^?    At  15^  ? 

1st  Solution.— At  6%  for  2|,  or  f  yr.,  the  interest  of  $20  is  $3; 
and  $30  +  $3  =  $33,  the  amount. 

3d  Solution,— The  interest  of  $1  for  2|  or  f  yr.,  at  6%,  is  15j^; 
and  $1  +  15^  =  $1.15,  the  amount.  If  $1  amounts  to  $1.15,  $20 
will  amount  to  30  times  $1.15  =  $33. 

The  interest  of  $1  for  f  yr.  at  9%  is  32^^ ;  and  the  amount,  $1.22^ ; 
hence  the  amount  of  $30  is  20  times  $1.33^  =  $34.50 ;  &c.,  &c. 

6.  What  is  the  amount  of  $30  for  3  yr.  3  mo.  at  8%  ? 
At  12^?    At  16^? 

7.  To  what  does  $50  amount  in  5  yr.  6  mo.,  at  6%  ? 
At  10^?    At  20^? 


Example  1.— What  is  the  interest  of  $250  for  4  yr.  at  6%  ? 

1st  solution. 

Explanation. — We  first  find  that 
part  of  the  principal  designated  by 
the  rate  for  1  yr.  to  be  $15,  by  mul- 
tiplying $350  by  .06 ;  we  then  multi- 
ply this  interest  for  1  yr.  by  4,  and 
find  the  interest  =  4  x  $15  =  $60.  Ans. 


Explanation. — We  first  mul- 
tiply the  interest  of  $1  for  1  yr. 
by  4  to  find  the  interest  of  $1 
for  4  yr.    This  is  equal  to  $.24. 

We  then  multiply  the  inter- 
est of  $1  for  4  yr.  by  350  to  find 
the  interest  of  $350,  which  is 
equal  to  $60,  Ans, 


$250, 
.06, 

Principal. 
Eate. 

$15.00, 
4 

$60 

Int.  for  1  yr. 
Int.  for  4  yr. 

$.06  = 
4 

2d  solution. 
Int.  of  $1  for  1  yr. 

$.24  = 
250, 

:  Int.  of  $1  for  4  yr. 
Principal. 

120 

48 

$60.00  =  Int.  of  $250  for  4  yr. 


326 


PE  RCEKTAGE, 


353.  EuLES. — I.  Multiply  the  principal  by  the  rate^  and 
that  product  hy  the  time.     {B  xr  x  t=F.)     Or, 

II.  Multiply  the  interest  of  $1  for  the  given  time  by  the 
nu7nber  of  dollars  in  the  principal.     (rxtxB=P.) 

As  the  amount  is  the  sum  of  the  principal  and  interest,  we  have 
only  to  add  these  parts  to  find  the  amount. 


Find  the  interest 

1.  Of  $150  for  2  yr.,  at  6^.  Ans.  $18. 

2.  Of  $275  for  3  yr.,  at  7^.  Ans.  $57.75. 

3.  Of  $128,121  for  8  yr.,  at  b%.  Ans.  $51.25. 

4.  Of  $650  for  5  yr.  at  %%.  Aiis.  $260. 

5.  Of  $225  for  8  mo.,  at  12^.  Ans.  $18. 

6.  Of  $580.75  for  6  mo.,  at  18^.  Ans.  $52.26f. 

Example  2. — What  is  the  amount  of  $750  for  3  yr.  at 


SOLUTION. 

$.08  =  Int.  of  $1  for  1  yr. 
3 

$.24  =  Int.  of  $1  for  3  yr. 
$1.00 


$1.24  =  Amt.  of  $1  for  3  yr. 
750,  Principal. 

620 
868 


Explanation. — We  first 
multiply  the  interest  of  $1 
fori  yr.  at  8%  by  3  to  find 
the  interest  for  3  yr.  To 
this  we  add  $1  to  find  the 
amount  of  $1  for  3  yr.,  at 
8%. 

This  sum  we  multiply 
by  750  to  find  the  amount 
of  $750  for  the  same  time 
and  at  the  same  rate.  There- 
fore, to  find  the  amount,  we 
have  this 


$930.00  =  Amt.  of  $750  for  3  yr. 

Rule. — III.  To  $1  add  its  interest  for  the  given  time^  at 
given  rate  ;  and  multiply  this  sum  by  the  number  of  dollars 
in  the  principal.    (1  -\- r  x  t  x  B  =  S.) 


INTEREST. 


327 


What  is  the  amount 

7.  Of  $150  for  2  yr.,  at  6%  ?  Ans.  I1G8. 

8.  Of  $275  for  3  yr.,  at  7%  ?  A)is.  $332.75.     . 

9.  Of  1756.25  for  2  mo.,  at  15^  ?  Ans.  ^775.16. 

10.  Of  $852.50  for  2.5  yr.,  at  9%  ?  Ans.  $1044.31. 

11.  Of  $116  for  8  yr.,  at  12^^  ?  Ans.  $232. 

12.  Of  $120.50  for  7  yr.,  at  15^  ?  Ans.  $247.03. 

13.  Of  $13520  for  3  yr.,  at  7^^  ?  Ans.  $16562. 

Example  3.— What  is  the  interest  of  $200  for  3  yr.  9  mo. 
18  da.,  at  7%  per  annum  ? 


Solution  by  Rule  I. 
$200,  Principal. 
.07,  Eate. 
$14.00,  Int.  for  1  yr. 
3.8,   Time  in  yr.  (340.) 
112 
42 


Solution  by  Rule  n. 
$.07  =  Int.  of  $1  for  1  yr. 
3.8      Time  in  yr. 


$.266  =  Int.  of  $1  for  3.8  yr. 

200 
$53.20,  Int.  of  $200  for  3.8  yr. 


$53.20,  Interest. 

What  is  the  interest 

14.  Of  $142.83  for  7  mo.  18  da.,  at  7%  ? 

15.  Of  $968.84  for  1  yr.  10  mo.  26  da.,  at 

16.  Of  $6360  for  2  mo.  17  da.,  at  4J^? 

What  is  the  amount 

17.  Of  $630.37i  for  9  mo.  15  da.,  at  5%? 

18.  Of  $2831.20  for  6  mo.  9  da.,  at  9%? 

19.  Of  $120  from  Jan.  1, 1874,  to  May  16,  1876,  at  8%? 
Note. — Find  time  between  dates  by  Rule,  p.  236. 

20.  Of  $360  from  Feb.  9, 1875,  to  May  18, 1876,  at  9%? 

Ans.  $401.31 


Ans.  $6.33. 
Ans.  $61.21. 

Ans.  $655.32. 
Ans.  $2964.97. 


328 


PERCENTAGE, 


21.  Of  $637.50  from  Mar.  4,  1873,  to  Aug.  16,  1876,  at 
10^?  Ans.  $857.43. 

22.  Of  $1184  from  Apr.  29, 1873,  to  Dec.  5,  1876,  at  6%  ? 

Ans.  $1397.12. 

23.  Of  $1580  from  May  20, 1872,  to  May  2, 1876,  at  4^  ? 

Ans.  $1829.64. 
What  is  the  interest 

24.  Of  $295.45  for  2  yr.  5  mo.  5  da.,  at  6^?    Ans.  S43.09. 

25.  Of  $238.38  from  Aug.  19,  1871,  to  Apr.  15,  1876,  at 
7^?  Ans.  $77.68. 

26.  Of  $487,375  from  June  15,  1875,  to  Sept.  30, 1875,  at 
1%  a  mouth  ?  Ans.  $17.06. 

27.  Of  $360  from  Aug.  10, 1876,  to  Nov.  20, 1876,  at  i% 
a  month  ?  Ans.  $9* 

Example  4. — What  is  the  interest  of  $150  for  5  yr.  9  mo. 
12  da.,  at  11%  per  annum  ? 

SoLtmoN  BY  Aliquots. 
$150  ,  Principal. 


.07  ,  Rate. 

$10.50  ,  Int.  for  1  yr. 
5 

ALIQUOTS. 

$62.50  ,  Int.  for  5  yr. 

6  mo.  =  ^  yr. 

5.25  ,    «      *'    6  mo. 

3  mo.  =  i  yr. 

2.625,    «      «    3  mo. 

12  da.  =^  yr. 

.35  ,    "      «  12  da. 

$60,725,  Int  for  5  yr.  9  mo.  12  da. 

Explanation. — ^We  first  find  the  interest  for  1  yr. ;  then  for  5  yr. 

For  6  mo.'s  interest  we  take  \  the  interest  for  1  yr. ;  for  3  mo.'s,  we 
take  \  the  interest  for  1  yr. ;  for  12  da,,  we  take  ^  the  interest  for  1 
yr. ;  and  the  sum  of  the  interest  for  5  yr.  +  ^  yr.  +  i  yr-  +  ^  yr. 
equals  the  interest  for  5  yr.  9  mo.  13  da.,  or  $60.73,  correct  An». 


INTEREST.  329 

What  is  the  interest 

28.  Of  $75  for  2  yr.  8  mo.  12  da.,  at  7^  ?      Ans,  114.18. 

29.  Of  $120  from  Jan.  1,  1873,  to  May  16,  1875,  at  S%  ? 

Ans.  122.80. 

30.  Of  $360  from  Feb.  9,  1875,  to  May  18,  1876,  at  9%? 

Ans,  $41.31. 

31.  Of  $637.50  from  Mar.  4,  1873,  to  Aug.  16,  1876,  a^ 
10^?  Ans.  $219.93. 

32.  Of  $1184  from  Apr.  29,  1873,  to  Dec.  5,  1876,  at  5%  ? 

Ans.  $213.12. 
Six  Per  Cent  Method. 

In  the  calculation  of  interest  we  reckon  only  360  days  in  a  year. 

If  then  the  interest  of  $1  for  1  yr.  is  6  cents,  or  60  mills,  the  interest 
for  1  day  must  be  ^^^  of  60  mills,  or  |  of  a  mill,  and  for  6  da.  it  must 
be  6  times  ^  of  a  mill,  or  1  mill ;  for  60  da.  or  2  mo.,  60  times  ^  of  a 
mill,  or  10  mills,  or  1  cent.    Therefore, 

If  the  interest  is  at  6%  per  annum,  the  interest  of  $1  is 
equal  to  one-half  as  many  cents  as  there  are  months,  or  ^  as 
many  mills  as  there  are  days. 

Having  found  the  interest  of  $1,  the  interest  of  any  number  of 
dollars  is  readily  found,  by  multiplying  this  interest  by  the  number  of 
dollars. 

And  having  found  the  interest  of  any  number  of  dollars,  at  6%,  we 
can  find  it  at  any  other  rate  %  hy  taking  such  a  part  of  that 
at  6fof  as  the  given  rate  %  is  of  Ofo, 

Example  5. — What  is  the  interest  of  $50,  for  18  mo. 
24  da.,  at  S%  ? 

SOLUTION. 

Int.  of  $1  for  18  mo.  ®  6%  =  ^^  or  $.09. 
Int.  of  $1  for  24  da.  @  6^  =  -^  m.,  or  $  004. 
Int.  of  $1  for  18  mo.  24  da.,  (^  6%  =  $.094. 

$.094  X  50  :=  $4.70,  Int.  of  $50,  @  6^. 

$4.70  X    f  =  $6.27,  Int.  of  $50,  @  8%- 


330  PERCENTAGE. 

What  is  the  interest  @  6% 

33.  Of  $200  for  200  mo.  ?  Ans.  $200. 

34.  Of  $150  for  18  mo.  12  da.  ?  Ans.  $13.80. 

35.  Of  $100  for  46  mo.  6  da.  ?  Ans.  $23.10. 

36.  Of  $80  for  62  mo.  15  da.  ?  Ans.  $25. 

37.  Of  $750  for  15  mo.  18  da.  ?  Ans.  $58.50. 

38.  Of  $425  for  4  yr.  4  mo.  9  da.  ?  Ans.  $111.14. 

39.  Of  $187.25  for  10  mo.  21  da.  ?  Ans.  $10.02. 

40.  Of  $250.63  for  9  mo.  27  da.  ?  Ans.  $12.40. 

41.  Of  $75  for  19  mo.  ?  Ans.  $7.13. 

What  is  the  amount 

42.  Of  $500  for  1  yr.  6  mo.  12da.,  at  8%  ?  Ans.  $561.33. 

43.  Of  $400  for  2  yr.  2  mo.  18  da.,  at  7^?  Ans.  $462.07. 

44.  Of  $200  for  2  yr.  6  mo.  24  da.,  at  6%  ?  Ans.  $225.67. 

45.  Of  $100  for  2  yr.  6  mo.  6  da.,  at  d%  ?     A7is.  $122.65. 

46.  Of  $150  for  3  yr.  8  mo.  9  da.,  at  10^  ?  Ans.  $205.38. 

47.  Of  $225  for  4  yr.  15  da.,  at  12^  ?  Ans.  $334.13. 

48.  Of  $350  for  4  yr.  21  da.,  at  4^  ?  ^/zs.  $406.81. 

49.  Of  $480  for  5  yr.  1  mo.  27  da.,  at  d%?  Ans.  $554.28. 

50.  Of  $645  for  6  yr.  3  mo.  20  da.,  at  7%?  Ans.  $929.69. 

Example  6.— What  is  the  interest  of  $600  for  33  da., 

at  e%? 

^  ( M.  ^  5t   No.  of  mills  interest  of  $1. 

SOLUTION.  \^    ^  J^^g  ^  g^^  ^  ^3  3^^  ^^^ 

What  is  the  interest 

51.  Of  $450  for  63  da.,  at  7%  ?  Ans.  $5.51. 

52.  Of  $525  for  93  da.,  at  S%  ?  Ans.  $10.85. 

53.  Of  $750  for  123  da.  at  9%?  Ans.  $23.06. 

54.  Of  $875  for  48  da.,  at  6%  ?  Ans.  $7. 


IlfTEREST.  331 

55.  Of  1900  for  78  da.,  at  10^?  Ans.  $19.50. 

56.  Of  $1000  for  33  da.,  at  12^  ?  Ans.  $11. 

57.  Of  $75  for  63  da.,  at  6%?  Ans.  $.79. 

58.  Of  $150  for  34  da.,  at  6%  ?  Ans.  $.71. 

59.  Of  $225  for  93  da.,  at  7^?  Ans.  $4.07. 

60.  Of  $540  for  123  da.,  at  8%  ?  Ans.  $14.76. 

All  of  our  calculations  in  Interest,  thus  far,  have  proceeded  on  the 
assumption  that  there  are  only  360  da.  in  1  yr.,  whereas  there  are  365. 

In  some  States,  calculations  of  interest  are  made  on  the  latter  basis. 
In  which  case  we  need  make  no  changes  so  far  as  entire  years  axe 
concerned ;  but  when  there  is  a  part  of  a  year  we  have  this 

Rule. — Calculate  the  interest  by  the  methods  already 
adopted,  and  subtract  from  the  result  -^^  of  itself. 

The  accuracy  of  this  Rule  is  evident  from  the  fact  that  865  da.  — 
360  da.  =  5  da,  ;  and  5  da.  are  ^f  5,  or  -^^^  of  a  year.  So  that  5,  or 
any  number  of  days  in  the  latter  method,  contains  tj^^  less  time  than 
the  same  number  of  days  in  the  former ;  and  as  a  consequence  the 
interest  should  be  ^^  less. 

In  accordance  with  this  Rule  find  the  true  interest  in 
Problems  51  and  60  inclusive. 

Example  7.— Find  the  interest  of  £45  12s.  6d.  for  3  yr. 
6  mo.  18  da.,  at  b%. 

Explanation. — We  first 
SOLUTION.  find  the  value  of  £45  12s. 

£45  12s.  6d.         =  £45.625  6d.  in  pounds  sterling. 

Rate .05  Multiplying  this  result  by 

the  given  rate  gives  the  in- 


Int.  for  1  yr.     ...        £2.28125  terest  for  1  year. 

Time  in  yr 3.55  Multiply  the  interest  for 

1  yr.  by  3.55,  the  time  in 

Int.  for  given  time  £8.0984375  =     ^ears,  and  we  have  the  in. 
£8  Is.  ll|d.,  Ans.    '  terest  =  £8.0984375  =  £8 

Is.  llfd.,  Ans. 

The  method  by  aliquot  parts  or  any  other  method  may  be  used. 


332 


PERCEI^TAGE. 


61.  What  is  the  interest  of  £50  6s.  9d.  for  4  yr.  1  mo. 
15  da.,  at  b%  ?  Ans,  £10  7s.  7d.  2.825  far. 

62.  What  is  the  amount  of  £75  4s.  6d.  for  5  yr.  2  mo. 
20  da.,  at  ^%  ?  Ans.  £92  18s.  .69d. 

63.  What  is  the  amount  of  £100  10s.  3d.  for  8  yr.,  at 
^%  ?  Ans.  £148  15s.  2.04d. 


CASE    II. 

354.  The  !Principal^  Interest,  and  Time  given, 
to  find  the  Mate  %. 


Example.— At  what  %  will  150  yield  $7,  in  2  yr.  4 
mo.? 

Solution. — 2  yr.  4  mo.  =  2^  yr.  If  $50  yield  $7  in  2|,  or  \  yr. ,  in 
1  yr.  it  will  yield  $3  ;  and  if  $50  yield  $3  in  1  yr.,  $1  will  yield  ^^  of 
$3,  or  6^.    Therefore,  $50  will  yield  $7  in  2  yr.  4  mo.,  at  6% .     Or, 

Since  the  interest  of  $1  for  2  yr.  4  mo.,  or  2^  yr.,  at  1^  is  2^,  or  \f, 
the  interest  of  $50  for  the  same  time  and  at  the  same  rate  is  50  times 
If  —  %\.  And  if  the  interest  at  1  %  yields  %%,  in  order  to  yield  $7, 
the  rate  %  must  he  as  many  as  $|  is  contained  times  in  $7,  or  6  times. 
Therefore,  $50  will  yield  $7  in  2  yr.  4  mo.,  at  6%. 


PB  OBLEM8. 

At  what  %  will 


$300  pay  $45  in  3  yr.  ? 
$450  pay  $135  in  5  yr.  ? 
$575  pay  $184  in  4  yr.? 
$625  pay  $50  in  2  yr.? 


1. 
2. 
3. 
4. 
9.  At  what  %  will  1100  pay  $10  in  6  mo.  10  da.  ? 


5.  $300  pay  $63  in3yr.? 

6.  $250  pay  $45  in  2  yr.  3  mo.  ? 

7.  $500  pay  $70  in  2  yr.  4  mo.? 

8.  $10  pay  $1.40  in  3  yr.  4  mo.  ? 


INTEREST.  333 
,4.^ 


^  IWieitteii  ^Aler  eisc 


Example.— At  what  %  will  $150  pay  $20.25  in  2  yr. 
3  mo.? 

Explanation. — We  first 

SOLUTION.  divide  the  given  interest 

$20.25  -r-  150  =  $.135  hy  the  given  principal  to 

($.135  ^  2})  X  100  =  6,  the  rate  %.     ^^^  *^«  ^^^e  for  2J  years. 

^  We  next  divide  this  rate 

'  by  2J  to  find  the  rate  for 

^^•^^       V  1 0O'/  —  fi^  1  y^-  =  06,  which  multi- 

150  X  2i  "^         /'  ~     /'•  pUed    by  100,  gives    the 

rate  % ,  6. 

355.  Rule. — Take  such  a  part  of  100%  as  the  given 
interest  is  of  the   product    of   the   principal  and  time. 

Divide  the  given  interest  hy  the  interest  of  the  given  prin- 
cipal for  the  given  time  at  1% ;  the  quotient  is  the  rate  %. 

Ult^B-"")    (354.^DS0Lnxi0^.) 

Note  the  introduction  of  the  new  element,  tiTne,  which  in  the  fop. 
mula  is  represented  by  "  t." 


PBOB  TjEMS. 

At  what  rate  %  will 

1.  $450  yield  $52.50,  in  2  yr.  4  mo.  ?  Ans.  5%, 

2.  $200  yield  $28.60,  in  2  yr.  4  mo.  18  da.  ?  Ans.  6%- 

3.  $150  yield  $60.72,  in  5  yr.  9  mo.  12  da.  ?  Ans.  7^. 

4.  $100  yield  $31.65,  in  3  yr.  6  mo.  6  da.?  ^ns.  9%. 

5.  $150  yield  $55,375,  in  3  yr.  8  mo.  9  da.?  Ans.  10%. 

6.  $225  yield  $131,625,  in  4  yr.  10  mo.  15  da.  ?  Ans.  12^. 


334 


PERCENTAGE. 


CASE     III. 

356.  TJie  Interest f  Tinie^  and  Rate 
to  find  the  JPHncipal, 


given, 


©l^at  ^^Xfoi  ci^fe  ^ 


Example.— What  principal  will,  in  3  yr.,  at  5%,  gain  $45  ? 


Solution. — Since  $1  in  3  yr.  at  5%  will  gain  $.15;  it  will  require 
as  many  dollars  to  gain  $45,  as  $.  15  is  contained  times  in  $45,  or  300 
times.     Therefore,  $300  will  gain  $45  in  3  yr.  at  5^. 


mOBIjEMS. 


What  principal  will  produce 


1.  $80  in  2  yr.  6  mo.,  at  8^  ? 

2.  $90  in  3  yr.  9  mo.,  at  4^  ? 

3.  $105  in  4  yr.  8  mo.,  at  6^? 

4.  $150  in  20  yr.,  at  b%  ? 

5.  $52.50  in  5  yr.  3  mo.,  at  h%  ? 

6.  $200  in  2  yr.  6  mo.,  at  S%  ? 

7.  $200  in  5  yr.,  at  5^? 


8.  $20  in  2  yr.,  at  b%  ? 

9.  $50  in  5  mo.,  at  1%%  ? 

10.  $100  in  2  yr.  6  mo.,  at  6^  ? 

11.  $75  in  10  yr.,  at  %^%  ? 

12.  $500  in  5  yr.,  at  20^? 

13.  $25  in  1  yr.  3  mo.,  at  12^  ? 

14.  $100  in  10  yr.,  at  10^  ? 


+lWi«itten  ^^Grci.8es"-6 


Example. — What  principal  will  in  3  yr.  4  mo.,  at 
produce  $100  ? 


1st  solution 
$100 


500. 


$.06  X  3f 
times.     Hence,  our  answer  is  $500. 


Explanation.  —  Since  $1  in  3^  yr.  at 
6%  will  produce  31^  times  6^  =  20/^,  it  will 
require  as  many  dollars  to  produce  $100 
as  20,«^  is  contained  times  in  $100,  or  500 


INTEREST.  335 

Explanation.— Interest  at  6%  for 
2d  solution.  31  yr.  =  20%,  or  i  of  the  principal. 

%      ^  *-,^^  _  dbKoo       ^^^^^  ^^^^  ^  ^  ^^  *^^  principal,  the 
6^  X  34  ~  ®^^^^®  principal  must  be  5  times  $100, 

or  $500. 

357.  Rules — I.  Divide  the  given  interest  ly  the  interest 
of  $1  for  the  given  time  at  the  given  rate  j  the  quotient  is 

the  principal     I =  ^.J 

II.  Take  such  apart  of  the  given  interest,  as  100%  is  of  the 
given  rate  %  multiplied  ly  the  given  time,    (d— ^  x  P  =  ^. ) 

What  principal  will  gain 

1.  $135  in  5  yr.,  at  6^  ?  Ans.  $450 

2.  $184  in  4  yr.,  at  8^  ?  Ans.  $575 

3.  $262.50  in  6  yr.,  at  ^%  ?  Ans,  $625 

4.  $400  in  4  yr.  2  mo.,  at  10^  ?  Ans.  $960 

5.  £252  in  3  yr.  6  mo.,  at  9^?  Ans.  £800 

6.  £50  in  2  yr.,  at  h%  ?  Ans.  £500 

CASE    IV. 

358.  The  Tinier  Hate  %,  and  Amount  given, 
to  find  the  JPrincipal* 


©Fat  ^:^GFci^G^ 


Example. — ^What  principal  will  in  1  yr.  8  mo.,  at  Q%, 

amount  to  $5.75  ? 

Solution. — Since  the  amount  of  $1  for  1  yr.  8  mo.,  at  9%  =  $1.15, 
it  will  require  as  many  dollars  to  amount  to  $5.75,  as  $1.15  is  contained 
times  in  $5.75,  or  5  times.  Therefore,  $5  will,  in  1  yr.  S  m«.,  at  9%, 
amount  to  $5.75. 


336 


PERCEIS^TAGE. 


PnOBT.  EMS. 


What  principal  will  amount 


1.  In  3    yr.,  at  Q%,  to  $118  ? 

2.  In  4J  yr.,  at  8^,  to  $68  ? 


3.  In  2^  yr.,  at  9^,  to  1363  r 

4.  In  5iyr.,atl0^,to$152? 


5.  In  1  yr.  9  mo.,  at  8^,  to  $228  ? 

6.  In  3  yr.  3  mo.,  at  12^,  to  $278  ? 

7.  In  6  yr.  2  mo.,  at  Q%,  to  $685  ? 

8.  In  1  yr.  8  mo.,  at  9^,  to  $230  ? 


+lWi«itten  ^T^ercise^fc 


Example.  —  What  principal  will  in  1  yr.  4  mo.,  at  6^, 
amount  to  $1188  ? 


SOLUTION. 

$1188 


$1  +  $.08 


$1100. 


Explanation. — Since  in  1  yr.  4  mo., 
at  6%,  $1  will  amount  to  $1.08,  it  will 
require  as  many  dollars  to  amount  to 
$1183  as  $1.08   is    contained    times   in 


$1188,  or  1100  times.     Therefore,  $1100  wHl  in  1  yr.  4  mo.,  at  6%, 
amount  to  $1188. 

359.   Rule. — Divide  the  given  amount  hy  the  amount  of 
$lfor  the  given  time  at  the  given  rate.    \-t^ T  ~  ^'1 


mOB  LEMS. 

What  principal  will  amount 

1.  In  6  yr.,  at  7^,  to  $887.50  ?  Ans.  $625. 

2.  In  2  yr.,  at  6%,  to  $784  ?  Ans.  $700. 

3.  In  3  yr.,  at  7^,  to  $166.37^? 

4.  In  2  yr.  4  mo.  18  da.,  at  6^,  to  $228.60  ?    Ans.  $200. 

5.  In  1  yr.  6  mo.  12  da.,  at  S%y  to  $561.33^  ?     Ans.  $500. 

6.  In  2  yr.  2  mo.  18  da.,  at  7^,  to  $462.06|?    Ans.  $400. 

7.  In  3  yr.  8  mo.  9  da.,  at  10^,  to  $205.37^?    Ans.  $150. 


INTEREST 


337 


CASE    V. 
360.  The  Interest,  JPrincipal^  and  Mate  % 
given,  to  find  the  Time. 


©tat  3E:^GFciXfeX 


Example. — In  what  time  will  $75  gain  $9,  at  6^  ? 

Solution.— At  G%,  $75  will  in  1  yr.  yield  75  x  $.06,  or  $4.50.  If 
$75  yield  $4|  in  1  yr.,  it  will  take  as  many  years  to  gain  $9  as  $4|  is 
contained  times  in  $9,  or  3  times.  Therefore,  it  will  take  $75,  at  6%, 
2  yr.  to  gain  $9. 

PROBLEMS. 

1.  In  what  time,  at  Q%,  will  $10  gain  $1.20  ? 
$1.80  ?     $2.40  ?    $7.20  ?     $8.70  ?     $10  ? 

Yc,  will  $15  gain  $1.05? 
$8.40?     $15? 
^,  will  $25  gain  $1.25  ? 
$4.33i? 


2.  In  what  time,  at  7 
$4.20?     $5.25?    $7.35? 

3.  In  what  time,  at  5 
$.93|  ?    $6.25  ? 


$1.50  ? 
$3.15? 


$2.50? 


$.62|? 


^lWl  itf  en  ^xerei^e^sT 


I- 


ExAMPLE^In  what  time  wiU  $137.50  gain  $28.87^,  at  7^? 

SOLUTION.  Explanation.  —  At  7%,   $137.50  will 

$28.87^  gain  in  1  yr.    137.50  x  $.07  =  $9,625;  and 

07  X  $137  50  ~  ^*       ^^  $137.50  gain   $9,625  in  1  yr.,  to  gain 

$38,875  will  take  as  many  yr.  as  $9,635  is 

contained  times  in  $38,875,  or  3  times.     Therefore,  at  7% ,  $137.50  will 

gain  $38,875  in  3  yr. 

361*  Rule. — Divide  the  given  interest  ly  the  interest  of 
the  given  principal  at  the  given  rate  for  one  year  ;  the  quo- 
tient is  the  time  in  years.     I 7^  =  t.\ 

22 


338  PERCENTAGE. 

fJROBLEMS  , 

In  what  time  will 

1.  1256.25,  at  5^,  yield  $51.25  ?  Ans,  4  yr. 

2.  '$450,  at  12^,  yield  $18  ?  Ans,  4^o. 

3.  $540,  at  8^,  yield  $14.88  ?  ^/is.  124  da. 

4.  $480,  at  3^,  yield  $74.28  ?       ^t^s.  5  yr.  1  mo.  27  da. 

5.  $225,  at  12^,  yield  $131,625  ?  Ans.  4  yr.  10  mo.  15  da. 

6.  $400,  at  7^,  yield  $62.06f  ?    Ans.  2  yr.  2  mo.  18  da. 

mOMISSOBY  NOTES. 

363.  A  JPromissory  Note  is  a  written  promise  to 
pay  a  specified  sum  of  money,  either  on  demand,  or  at  some 
designated  time  after  the  date  of  the  note. 

363.  When  the  time  of  payment  is  specified  in  the  note 
it  is  customary  to  allow  three  more  days,  called  Days  of 
Grace. 

364.  A  note  is  said  to  mature  on  the  day  on  which  it 
is  due. 

365.  The  promise  is  usually  either  to  pay  the  bearer, 
or  some  person,  mentioned  in  the  note,  or  to  Tiis  order, 

366.  A  Negotiable  Note  is  one  that  can  be  trans- 
ferred, or  sold. 

367.  A  Joint  Note  is  one  signed  by  two  or  more 
persons. 

368.  The  Face  of  a  Note  is  the  sum  promised  in  the 
note.  This  sum  should  always  be  written  in  words  in  the 
body  of  the  note. 

A  promissory  note  should  have  in  it  the  words  "  value  received,"  or 
"for  value  received,"  as  without  these  words,  the  holder  may  be 
required  to  prove  that  value  was  received. 

In  some  States  it  is  customary  also  to  insert  **  without  defalcation/* 
though  for  what  object  is  not  very  plain. 


INTEREST.  339 

369.  The  Maker,  or  Drawer,  of  a  note  is  the  per- 
son who  signs  it. 

370.  The  Holder,  or  Payee,  is  the  person  to  whom 
it  is  to  be  paid. 

371.  An  Indorser  is  the  person  who  writes  his  name 
upon  a  note,  or  other  obligation,  usually  upon  its  back, 
thereby  becoming  responsible  for  its  payment. 

37^.  A  JProtest  is  a  legal  notice  to  the  indorser  of  a 
note  that  the  maker  has  failed  to  pay  the  note  when  due. 

As  notes  are  not  always  paid  when  due,  and  holders  agree  to  take 
a  portion  of  the  payment,  and  wait  for  the  balance,  receiving  interest 
on  the  unpaid  part,  we  have  the  term 

373.  A  JPartial  JPayment,  which  is  a  payment  of 
part  of  a  note,  or  any  other  legal  obligation,  such  as  a  mort- 
gage^ article  of  agreement,  &c.  If  stated  on  the  back  of  the  in- 
strument in  writing  by  the  holder,  it  is  called  an  indorse^ 
meiit. 

JPABTIAL   PAYMENTS. 

To  compute  interest  when  partial  payments 
have  been  made. 

Settlements  are  made  by  the  Courts  of  the  United  States, 
and  by  most  of  the  individual  States,  by  what  is  known  as 
the  United  States  Rule. 

Example. 

$3000.  Pittsburgh,  Sept.  1, 1874 

On  demand,  for  value  received,  I  promise  to  pay  to  the 
order  of  J.  M.  Logan,  three  thousand  dollars,  without  defal- 
cation, with  interest  from  date. 

James  Meredith. 


340  PERCEN^TAGE. 

On  this  note  were  these  indorsements: 

Received,  Jan.  1/75.  $500  ;  March  1/75,  $25 ;  Apr.  7/75,  $600;  Oct. 
13/75,  $1000  ;  May  19/76,  $200.    How  much  was  due  June  1/76  ? 

SOLUTION. 

Principal $3000 

Int.  from  Sept.  1/74,  to  Jan.  1/75,  4  mo 60 

Amt.  of  Prin.  to  date  of  1st  Payment $8060 

1st  Payment  made  Jan.  1/75 500 

2d  Principal $25^0 

Int.  of  2d  Prin.  from  Jan.  1  to  Apr.  7/75 40.96 

Amt.  to  Apr.  7/75 $2600.96 

2d  Payment  (not  equal  to  Int.)  and  3d  Payment . . .       625. 

3d  Principal $1975.96 

Int.  of  3d  Prin.  from  Apr.  7  to  Oct.  13/75 61.255 

Amt.  of  3d  Prin.  to  Oct.  13/76 $2037.215 

4th  Payment 1000. 

4th  Principal $1037.215 

Int.  from  Oct.  13/75  to  May  19/76 37.340 

Amt.  of  4th  Prin.  to  May  19/76 $1074.555 

5th  Payment 200. 

5th  Principal $  874.555 

Int.  from  May  19  to  June  1/76 1.749 

Sum  due  June  1/70 $876,304 

374.  Itule  of  the  United  States  Courts. 

Find  the  amount  of  the  'principal  up  to  the  time  of  the 
first  payment,  if  that  payment  equals  or  exceeds  the  interest ; 
hut  if  it  does  not,  then  find  the  amount  up  to  the  time  when 
the  sum  of  two  or  more  payments  equals  or  exceeds  the 
interest. 

From  this  amount  subtract  the  first  payment,  or,  if  neces- 
sary, the  sum  of  the  two  or  more  payments. 

Use  the  remainder  as  a  new  principal,  and  proceed  as 
before,  until  time  of  final  settlement. 


INTEREST.  341 

PROBIjEMS. 

1.  $2500.  Philadelphia,  Feb.  12, 1873. 
On  demand  we  jointly  and  severally  promise  to  pay  Robert 

Mackay,  or  order,  twenty-five  hundred  dollars,  for  value 
received,  with  interest  from  date. 

Benjamin  Steinman. 

Lewis  A.  Morton. 

Indorsements.— June  12,  1873,  $25;  Dec.  12,  1873,  $150;  June 
12, 1874,  $750  ;  Dec.  12,  1874,  $350  ;  Apr.  12, 1875,  $250. 

What  was  due  Apr.  12,  1876?  Ans.  $1331.61. 

2.  $7500.  New  Orleans,  July  1,  1874. 
Six  months  after  date,  I  promise  to  pay  to  the  order  of 

Samuel  Hill  and  Co.,  seven  thousand  five  hundred  dollars,  at 

the  First  National  Bank  of  New  Orleans,  for  value  received. 

Due  Jan.  1/4,  1875.  B.  Simpson. 

Indorsements.— Jan.  4, 1875,  $150;  July  4,  1875,  $1000;  Aug.  19, 
1875,  $1000 ;  Sept.  4. 1875,  $3000. 

Find  balance  due  Jan.  4,  1876,  at  h%.      Ans.  $2629.30. 

3.  $1500.  New  York,  Aug.  26,  1873. 
Four  months  after  date  I  promise  to  pay  J.  Bos  well,  or 

order,  fifteen  hundred  dollars,  for  value  received. 
Due  Dec.  26/29, 1873.  John  Tweddle. 

Indorsements.— June  29, 1874,  $50  ;  Dec.  29, 1874,  $350  ;  Mar.  29, 
1875,  $20 ;  Sept.  29, 1875,  $370  ;  Dec.  29, 1875,  $200. 

What  was  due  June  29,  1876,  at  1%  ?        Ans.  $717.91. 

4.  $1750^.  St.  Louis,  Oct.  25, 1874. 
Four  months  after  date,  I  promise  to  pay  to  the  order  of 

J.  B.  Jones,  one  thousand  seven  hundred  fifty  dollars  and 
twenty-five  cents,  at  the  Merchants'  Bank,  value  received. 
Due  Feb.  25/28,  1875.  G.  J.  Luckey. 


342  PERCEN^TAGE. 

Credits  on  this  Note.— Feb.  28,  1875,  $50  25;  June  14,  1875, 
$25  ;  Oct.  20,  1875,  $350 ;  May  8,  1876,  $575. 

Find  balance  due  Sept.  8,  1876,  at  Q%.       A^is.  $878.86. 
Many  business  men  settle  notes  and  accounts  on  short 
time  according  to  the  following,  often  called 

375.  The  Commercial  or  Merchants^  Hale, 

Find  the  amount  of  the  principal  at  the  time  of  settlement. 
Then  find  the  amount  of  each  payment  from  the  time  it  loas 
made  until  settlement,  and  suhtract  the  sum  of  the  amounts 
of  the  payments  from  the  amount  of  the  principal. 

Example. 

$950.  Columbus,  0.,  Feb.  3, 1876. 

On  demand,  I  promise  to  pay  to  the  order  of  H.  I. 

Gourley,  nine  hundred  fifty  dollars,  without   defalcation, 

value  received.  Edwaed  M.  JoHNSTOiq^. 

Indorsements.— Mar.  1,  '76,  $150 ;  June  3,  '76,  $96 ;  July  8,  76, 
$300  ;  Dec.  20,  '76,  $250. 

How  much  was  due  Jan.  17,  1877? 

SOLUTION. 

Amount  of  $950,  from  Feb.  3,  '76,  to  Jan.  17,  '77,  $1004466f. 

Amount  of  $150,  from  Mar.  1,  '76,  to    "      "     "  $157.90 

Amount  of  $  96,  from  June  3,  '76,  to    "     "     *'  $  99.584 

Amount  of  $300,  from  July  8,  76,  to    "      *'     "   $309.45 

Amount  of  $250,  from  Dec.  20,  '76,  to    "     "     "   $251,125  i  818.059. 


Am.  $  186.41. 
The  following  notes  bear  interest  from  date : 
6.  Face  $1000;  date  Apr.  16,  1876. 
Payments.— $400,  Aug.  16, 1876  ;  $360,  Oct.  20, 1876. 
Balance  due  Feb.  25,  1877,  at  6^  ?  Ans.  $271.40. 

6.  Face,  $1500 ;  date,  Aug.  16,  '75  ;  rate  %,  10. 


INTEREST.  343 

Payments.— $300,  Nov.  1, 75 ;  $200,  Feb.  26, 76 ;  $250,  Apr.  22, 76 ; 
$400,  June  1,  76. 

Balance  due  Aug.  16,  1876  ?  Ans,  $450.56. 

376.  In  Connecticut,  the  Supreme  Court  has  adopted 
the  following  principles  in  the  calculation  of  interest : 

First. — Payments  made  when  Merest  has  run  a  year  or 
more,  and  those  less  than  the  interest,  are  treated  as  in  the 
U.  S.  rule. 

Secoi^d. — A  payment  made  within  a  year  from  the  begin- 
ning of  any  interest,  draws  interest  for  the  rest  of  that  year, 
if  that  year  does  not  extend  heyond  settlement;  and  its 
amount  must  ie  taken  from  the  aynount  of  the  principal  for 
that  year.  But  if  the  year  does  extend  leyond  settlement, 
the  amounts  are  computed  for  hotJi  principal  and  payment, 
to  settlement.  The  difference  of  these  amounts  is  the  balance 
due. 

In  some  States,  as  in  Vermont  and  New  Hampshire,  what 
is  called  ^^ Annual  Interest,'^  is  allowed;  that  is,  if 
interest  is  not  paid  when  due,  it  will  bear  simple  interest. 

Example. — $100  is  to  be  paid  in  3  yr.  with  annual  interest.  At  the 
end  of  the  1st  year  $6  int.  is  due.  If  these  $6  are  not  paid  until  the 
end  of  the  three  years,  two  years  simple  interest  ($6  x  .13  =  $.72) 
must  be  added  to  the  interest.  In  like  manner,  if  the  $6  int.  due  at 
the  end  of  the  second  yr.  is  not  paid  until  the  end  of  the  3d  year,  one 
year's  simple  interest  ($6  x  .06  =  $.36)  must  be  added.  This  will 
make  the  sum  paid  at  the  end  of  the  three  years  =  Principal,  $100  + 
Ist  year's  int.,  $6  +  2yr.'s  int.  on  1st  year's  int.,  $.72  +  2d  yr.'s  int., 
$6  +  1  yr.'s  int.  on  2d  yr.'s  int.,  $.36  +  3d  yr.'s  int.,  $6  =  $119.08, 
or  $1.08  more  than  by  TJ.  S.  rule. 

COMPOUND   IlfTEBEST. 

In  Compound  Interest,  as  in  Simple  Interest,  there  are  5  cases.  It 
is,  howerver,  thought  best  to  discuss  the  one  following,  only  : 


344  PERCE  KTAGE. 

377.  TJie  rrincipal,  Mate  %  and  Time  given, 
to  find  the  Amount,  and  Interest, 

Example.— Find  the  amount  and  interest  of  $200  for 
2  yr.  3  mo.,  at  5^. 

SOLUTION. 

Principal $200. 

Amt.  at  end  of  Ist  year,  $200       x  1.05    --  $210. 
,  Amt.  at  end  of  2d  year,  $210       x  1.05    =  $220.50 

Amt.  at  end  of  2^  years,  $220.50  x  1.01^  =  $223.26 

Subtract  the  Principal $200.00 

And  there  remains  the  interest $  23.26 

378.  Rule. — Find  tlie  amount  for  the  1st  period  of  time, 
and  malce  it  the  principal  for  the  2d  period;  mahe  the 
amount  of  the  2d  the  principal  for  the  3d,  and  so  on.  The 
last  amount  will  he  the  one  sought. 

SuUract  the  1st  Principal  from  the  last  amount ;  the 
remainder  is  the  compound  interest. 

T  n  OB  L  EMS. 

At  compound  interest  find  the  amount  and  interest 

1.  Of  $300  for  2  yr.  4  mo.,  at  6%.     Ans.  $343.82;  $43.82. 

2.  Of  $500  for  5  yr.,  at  7^.  Ans.  $701.27;  $201.27. 

3.  Of  $350  for  4  yr.  6  mo.,  at  3^.    Ans.  $399.84;  $49.84. 

4.  Of  $600  for  7  yr.,  at  8^.  Ans.  $1028.29;  $428.29. 

5.  Of  $125  for  8  yi\,  at  4^.  Ans.  $171.07;  $46.07. 

6.  Of  $375  for  6  yr.,  at  5^.  Ans.  $502.54  ;  $127.54. 

Note.— Observe  that  the  amount  of  any  principal  for  a  given  num- 
ber of  years  is  equal  to  the  product  of  1  plus  the  rate  used  as  a  factor 
as  many  times  as  there  are  units  in  the  number  of  years,  multiplied  hy 
the  principal.  Thus,  Ex.  2  is,  1.07  x  1.07  x  1.07  x  1.07  x  1.07  x  $500 
=  $701.27  +  . 


II^TEREST. 


345 


The  labor  of  computing  compound  interest  is  greatly  lessened  by 
the  use  of  the  following 

379.    TABLE, 

Showing  the  amount  of  1  unit  of  money  at  Compound  Interest,  at 
from  dfo  to  8%  for  periods  not  greater  than  35  yr. 


Yr. 
1 

3  per  cent. 

4  per  cent. 

5  per  cent. 

6  per  cent. 

7  per  cent. 

8  per  cent. 

1.03 

1.04 

1.05 

1.06 

1.07 

1.08 

2 

1.0609 

1.0816 

1.1025 

1.1236 

1.1449 

1.1664 

3 

1.092727 

1.124864 

1.157625 

1.191016 

1.225043 

1.259712 

4 

1.125509 

1.169859 

1.215506 

1.262477 

1.310796 

1.360488 

5 

1.159274 

1.216653 

1.276282 

1.338226 

1.402551 

1.469328 

6 

1.194052 

1.965.319 

1.340096 

1.418519 

1.500730 

1.586874 

7 

1.229874 

1.315932 

1.407100 

1.503630 

1.605781 

1.713824 

8 

1.266770 

1.368569 

1.477455 

1.593848 

1.718186 

1.850930 

9 

1.304773 

1.423312 

1.551328 

1.689479 

1.838459 

1.999004 

10 

1.343916 

1.480244 

1.628895 

1.790848 

1.967151 

2.158924 

11 

1.384234 

1.539454 

1.710339 

1.898299 

2.104851 

2.a31638 

12 

1.425761 

1.601032 

1.795856 

2.012196 

2.252191 

2.518170 

13 

1.468534 

1.665074 

1.885649 

2.13i928 

2.409845 

2.719623 

14 

1.512590 

1.731676 

1.979932 

2.260904 

2.578534 

2.937193 

15 

1.557967 

1.800944 

2.078.928 

2.396558 

2.759031 

3.172169 

16 

1.604706 

1.872981 

2.182875 

2  540352 

2.952163 

3.425942 

17 

1.652848 

1.947901 

2.292018 

2.69*773 

3.158815 

3.700013 

18 

1.70M33 

2.025817 

2.406619 

2.854a39 

3.379932 

3.996019 

19 

1.753506 

2.106849 

2.526950 

3.025600 

3.616527 

4.315701 

20 

1.806111 

2.191123 

2.653298 

3.207135 

3.8696&4 

4.660957 

21 

1.860295 

2.278768 

2.785963 

3.399564 

4.140562 

5.033833 

22 

1.916103 

2.369919 

2.925261 

3.603537 

4.430401 

5.4.36540 

23 

1.973587 

2.464716 

3.071324 

3,819750 

4.740529 

5.871463 

24 

2.032794 

2.56:3304 

3.22510'J 

4.0489:35 

5.072366 

6.341180 

25 

2.093778 

2.665836 

3.386355 

4.301871 

5.427432 

6.848475 

Example  2. — What  is  the  amount  and  interest  of  S250 
for  9  yr.  6  mo.,  at  7%  ? 

SOLUTION. 

Ami  of  $1  for  9  yr.,  @  7^  =  $1.838459 

250 

Amt.  of  $250  for  9  yr.,  @  7^  =  1459.615 
Amt.  of  $1  for  6  mo.,  @  7%  =  1.03^- 


Amt.  of  $250  for  9  yr.,  6  mo.  %1%  =    $475.70 
Principal $250 


Int.  of  $250  for  9  yr.  6  mo.  %1%  =    $225.70 
Explanation. —We  first  find  in  the  Table  opposite  9  in  the  1st 


346  PERCENTAGE. 

column,  and  under  the  column  headed  7%,  that  the  amount  of  $1  for 
9  yr.  @  7%  is  $1.838459.  Multiplying  this  by  250,  we  find  the  amount 
of  $250  for  the  same  period  =  $459,615.  This  result  multiplied  by 
the  amount  of  $1  for  6  mo.,  at  7%,  gives  us  $475.70,  the  amount  of 
$250  for  9  yr.  ^  mo.,  at  7%. 

Subtracting  the  original  principal  from  the  last  amt.  leaves  the 
interest  =  $225.70. 

380.  Rule. — For  a  whole  number  of  years.  Multiply 
the  amount  of  one  unit  of  money  hy  the  number  of  such  units 
in  the  principal.  The  result  is  the  amount.  From  this  sub- 
tract the  principal     The  result  is  the  compound  interest. 

For  a  final  fraction  of  a  year.  Use  the  amount  of  the 
whole  number  of  years  as  a  principal  for  the  fractional  time. 

PROBLEMS, 

1.  What  is  the  compound  interest  of  $300  for  18  yr.  9  mo., 
at  4^,  payable  annually?  Ans.  $325.98. 

2.  What  is  the  compound  interest  of  $400  for  20  yr.  4  mo. 
;24  da.,  at  b%,  payable  annually  ?  Ans.  $682,55. 

3.  What  is  the  compound  interest  of  $1000  for  25  yr. 
8  mo.,  at  6^,  payable  annually  ?  Ans.  $3463.55. 

4.  What  is  the  interest  of  $1600  for  8  yr.  6  mo.,  at  6^, 
payable  semi-annually  ?  Ans.  $1044.56. 

Note.— Observe  that  this  is  the  same  as  17  yr.,  at  3%. 

5.  What  is  the  interest  of  $1000  for  8  yr.  6  mo.,  at  8^, 
payable  semi-annually  ?  Ans.  $947.90. 

6.  What  is  the  interest  of  $100  for  42  yr.,  at  S%  ? 

Ans.  $2433.94. 

Note. — First  find  amount  for  25  yr. ;  then  using  this  amount  as  a 
new  principal,  find  its  amount  for  17  yr. 


SECTION    VII 


><S)T(j£T^^ 


5^ 


e> 


IDIS3COWMT' 


<?> 


-^ 


^*<^£(£>^ 


381.    Discount  is  a  deduction  from  a  price,  or  from 

a  debt. 

Discount  is  of  three  kinds,  namely:  Trade,  Bank,  and  True  or 
Equitable,  Discount. 

383.  Trade  Discount  is  a  deduction  from  a  price, 
or  from  cost.  Thus,  a  seller  is  said  to  discount  7iis  goods, 
when  he  deducts  a  part  of  the  price.  An  article  is  some- 
times said  to  be  sold  at  a  discount,  when  it  is  sold  for  less 
than  it  cost. 

Trade  Discount  is  usually  reckoned  as  a  percentage  of  the  price, 
or  cost. 

383.  IBank  Discount  is  a  deduction  made  from  the 
face  or  amount  of  a  note,  or  other  obligation,  due  at  some 
future  time. 

384.  A  Hank  is  an  institution  legally  established  for 
the  purpose  of  deahng  in  money. 

385.  A  Bank  of  Deposit  is  a  bank  legally  em- 
powered to  receive  and  take  charge  of  the  money  of  others. 

386.  A  Deposit  is  money,  or  its  equivalent,  intrusted 
to  the  keeping  of  a  bank. 

387.  A  Depositor  is  a  person  who  makes  a  deposit. 

388.  A  Check  is  a  written  order  on  a  bank  for  money. 

389.  A  JBanIc  of  Issue  is  a  bank  legally  empowered 
to  pay  out  its  own  promissory  notes  as  money.  These  notes 
are  frequently  called  bank  notes,  or  bank  bills. 

^47 


348  PEKCEKTAGE. 

390.  A  Sank  of  Discount  is  a  bank  legally  em- 
powered to  loan  money. 

Banks  of  Deposit,  of  Issue,  and  of  Discount  are  often  combined  in 
one  institution. 

The  affairs  of  a  bank  are  usually  managed  by  a  Board  of  Directors, 
chosen  by  the  Stockholders. 

The  board  of  directors  select 

A  President,  who  presides  over  their  meetings,  and  in  a  bank  of 
issue,  signs  all  bills,  or  notes  issued  ; 

A  Cashier,  who  has  charge  of  the  accounts  of  the  bank,  and  who, 
also,  signs  bills,  or  notes ; 

One  or  more  Tellers,  who  are  called  receiving  and  paying  tellers,  to 
designate  their  position. 

A  bank  is  said  to  discount  a  note  when  it  purchases  the  note  for  a 
sum  less  than  the  note  will  be  worth  when  it  becomes  due. 

The  sum  which  is  paid  for  a  note  is  called  the  avails,  or  proceeds,  of 
the  note. 

In  bank  discount  three  days  of  grace  are  always  considered  as  part 
of  the  time  a  note  has  to  run.  Hence,  a  note  for  30  days  is  always 
considered  33  days  ;  one  for  60  days,  63  days,  &c. ;  and  the  discount  is 
reckoned  also,  for  this  additional  time, 

391.  True^  or  Equitable  jyiseoiint^  is  also  a  de- 
duction made  from  the  face  or  amount  of  a  note,  or  other 
obligation  due  at  some  future  time ;  but  it  differs  from  Bank 
Discount  in  this  respect, — 

393.  In  Bank  Discount,  the  face  of  the  note,  if 
luitJiout  interest,  or  the  amount  of  the  face,  if  with  interest, 
is  taken  as  the  base  of  calculation,  and  the  discount  is  calcu- 
lated exactly  as  Simple  Interest,  Case  I ;  while  in  True 
Discou7it,  not  the  face  of  the  note,  but  a  sum  which, 
in  a  given  time,  will  amount  to  the  face  or  amount  of  the 
note,  is  the  base. 

For  Example.— The  bank  discount  of  $106  for  1  yr.,  at  6%  (no 
days  of  grace)  =  $6.36 ;  but  the  true  discount  of  $106  is  -^  of  $100 
=  $6  :  for  $100  is  the  sum  which  in  1  yr.,  at  6% ,  will  amount  to  $106. 


DISCOUNT.  349 

Bank  Discount. 

Bank  Discount  has  Five  Cases,  which  correspond  to  the 
Five  Cases  in  Simple  Interest. 

CASE    I. 

393.  Tlie  Bate  %,  Time,  and  Face  or  Amount 
of  Note  given,  to  find  the  Discount. 

Example. — What  is  the  bank  discount  of  $350,  due  3  mo. 
hence,  at  %%  discount  ?    What  are  the  proceeds  ? 

SOLUTION  Explanation.— Adding 

90  da.  +  3  da.  =  93  da..  Time.  I  ^^  ^'^'^'^.^  ^^^  ^^ 

*  ^  da.,  time  of  Dis.  =  AV  or 

$350  X  .08  X  tV.  =  i7.23J,  Dis.        ^a^/  y^.      The   interest  of 

$350  —  $7.23i  =  $342.77,  Pro.  $350,  the  face  of  the  note, 

for  xV?r  yr»  at  8 fc  =  $7.23^, 
or  discount.     The  proceeds  =  $350  -  $7.33^  =  $342.77. 

The  avails,  or  proceeds,  may  be  found  by  subtracting  the  discoumt 
from  the  face,  or  amount,  of  the  note. 

394.  KuLES. — Compute  the  interest  on  the  face  of  the 
note  {or,  if  it  hears  interest,  on  its  amount  when  due)  at 
the  given  rate  %  for  three  days  more  than  the  given  time  ; 
the  result  is  the  bank  discount.     (B  x  r  x  t  =  P.)     Or, 

Multiply  the  interest  of  Si,  for  the  given  time,  at  given  rate, 
ly  the  numher  of  dollars  to  he  discounted,    (r  xt  x  B=  P.) 

PROBLEMS. 

1.  What  is  the  discount  of  $250,  discounted  for  4  mo., 
at  9%  ?    What  are  the  proceeds  ?      Ans.  $7.69  ;  $242.31. 

2.  What  are  the  discount  and  proceeds  of  a  note  for  $75Q 
discounted  for  90  da.,  at  S%  ?  Ans.  $15.50 ;  $734.50. 


35Q  PERCENTAGE. 

3.  What  are  the  discount  and  avails  of  a  note  for  $3250, 
discounted  for  60  da.,  at  1%  ?         A7is.  139.81 ;  $3210.19. 

4.  A  note  for  $650  dated  Jan.  1/76,  due  in  3  mo.,  was 
discounted  Mar.  1/76,  at  Q%.  What  were  the  discount  and 
proceeds?  ^ns.  $3.68  ;  $646.32. 

Note. — When  a  note  is  made  for  a  given  number  of  months,  its 
maturity  is  reckoned  by  calendar  months,  but  its  Discount  by  the 
actual  number  of  days. 

CASE    II. 

395.  The  Discount^  Time,  and  Face,  or 
Amount  given,   to  find  the  Hate  %, 

Example. — The  discount  of  a  3  mo.  note  for  $100,  was 
$1.55.    What  was  the  rate  %  discount  ? 

SOLUTION.  Explanation. — The  int. 

90  da.  +  3  da.  =  93  da.  ^^  ^1^^  ^«^  ^^  ^^^  «^  xWyr-, 

$100  X  .01  X  AV  =  ^-2581,  Dis.       ^!  1^^  ''I  ^-^.^^^  =  discount. 
*-^^        *.^^r.:         «  /.^  Smce  the  ffiven  discount  is' 

$1.55  -  $.3584  =  6,  or  6^.  ^^  ,._  ^^^  ^^^  ^^  ^^^^  ^ 

as  many  times  1  ^  as  $.258^ 
are  contained  times  in  $1.55,  or  6  times,  or  6%. 

396.  KuLES. — Divide  the  given  discount  hy  the  discount 
of  the  face  {or  amount)  for  the  time  at  1%.     The  quotient 

is  the  rate  %.    (^--^^  =  e)    Or, 

Take  such  U  part  of  100%  as  the  given  discount  is  of 
the  product  of  the  time  and  face  (or  amount)  of  the  note. 


(_^^XiO.^=iJ.) 


PBOBLEMS. 

Find  the  rate  %  at  which  a  note  is  discounted  when 
1.  The  face  is  $400 ;  discount,  $4.20 ;  time,  60  da. 

Ans.  6^ 


DISCOUNT.  351 

2.  The  face  is  $150 ;  discount,  $4.10  ;  time,  4  mo.     What 
is  the  rate  %  ?  Ans.  8%. 

3.  The  face  is  $250 ;  discount,  $5.81 ;  time,  3  mo.    What 
rate  %  is  charged  ?  Ans.  9^. 

4.  The  face  is  $800  ;  discount,  $14.47  ;  time,  3  mo.     Re- 
quired the  rate  %.  Ans.  7%- 

5.  The  face  is  $900;  discount,  $19.20 ;  time,  61  da.    What 
is  the  rate  %  of  discount  ?  Ans.  12^. 

CASE    III. 

397.  The  Discount,  Time^  and  Hate  %  given, 
to  find  the  Face. 

Example. — The  discount  of  a  4  mo.  note,  at  8%,  was 
$4.10.     What  was  the  face  ? 

Explanation.  —  The  int. 

SOLUTION.  or  discount  of  $1  for  123  da., 

120  da.  4-  3  da.  =  123  da.  or  ^  jr.,  at  8%  is  $.0273^. 

II  X  .08  X  tV^t  =  $.0273^,  Dis.      Since  the  given  dis.  is  $4.10, 

$4.10  -  $.0273J  =  150,  or  *^"  ^^"^  ^/.  *^"  ^f "  ^^,to? 

41  KA    T?  ^®  many  times  $1  as  $.0273^ 

$150,  h  ace.  ^^^  contained  times  in  $4.10, 

or  150  times,  or  $150,  Face. 

398.  EuLE. — Divide  the  given  discount  hy  the  discount 
of$l  for  the  given  time,  at  given  rate  ;  the  quotient  is  the  face, 

\r  x  t  I 

ntOBLEMS  . 

1.  Discount  is  $11.96 ;  time,  4  mo. ;  rate  %,  10.    What  is 
the  face  of  note  ?  Ans.  $350. 

2.  Discount  is  $2.75 ;  time,  1  mo. ;  rate  %,  6.    Find  face 
of  the  note.  Ans.  $500. 


352  PEKCEKTAGE. 

3.  Discount  is  $4.20;  time,  60  da.;  rate  %,  6.     Eequired 
the  face  of  note.  Ans.  $400. 

4.  Discount  is  $14.47 ;  time,  3  mo. ;  rate  %,  7.    Face  of 
note  required.  Ans,  1800. 

5.  Discount  is  $13.06 ;  time,  91  da. ;  rate  %,  5.    What  is 
the  face  of  note  ?  Ans,  $1000. 

CASE    lY. 

399.  The  Tinie^  Mate  %^  and  Proceeds  given^ 
to  find  the  Face, 

Example. — What  must  be  the  face  of  a  note  which,  dis- 
counted for  4  mo.,  at  8^,  will  yield  $145.90  ? 

SOLUTION.  Explanation. — The  int. 

130  da.  +  3  da.  =  123  da.  o^  dis.  of  $1  for  123  da.,  or 

$1   X  .08  X  ^  =  $.02731,  Dis.       ^Viy   yr.,  at  8%  is   $.0273,^. 
$1  -  $.0273i  =  $.97261,  f>ro.  ^.^0273^^2;,  lincf tlTe 

$145.90  -  $.9726f  =  150,  or  ^^,^  proceeds  are  $145.90, 

$150,  Face.  the  face  of  the  note  must  be 

as  many  times  $1  as  $.9726| 
are  contained  times  in  $145.90,  or  150  times,  or  $150,  Face. 

400.  KuLE. — Divide  the  given  proceeds  hy  the  proceeds 
of  $1  for  the  time,  at  rate  given ;  the  quotient  is  the  face 

of  the  note.    (^ — 7 -r  =  B.) 

PBOBZEMS, 

What  must  be  the  face  of  a  note  payable  in 

1.  Thirty  days,  rate  %,'6,  to  draw  $100  ?      Ans.  $100.55. 

2.  Ninety  days,  rate  %,  6,  to  draw  $295.35  ?     Ans.  $300. 

3.  Three  months,  rate  %,  9,  to  draw  $244.19  ?  Ans.  $250. 

4.  Ninety  days,  rate  %y  15,  to  draw  $1000  ? 

Ans.  $1040.31. 


DISCOUNT.  353 

5.  Time,  four  months;  rate  %,  8;  proceeds,  $2165.45. 
What  is  the  face  of  note  ?  Ans.  $2226.30. 

6.  Time,  four  months ;  rate  %,  8 ;  proceeds,  $7350.    Re- 
quired the  face  of  note.  Ans.  $7556.54. 

7.  Time,  five  months ;  rate  %,  10  ;  proceeds,  $800.     Face 
of  note  required.  ^?^5.  $835.51. 

CASE    V. 

401.    The   Face,  Mate  %  and  Discount    or 
Proceeds  given,  to  find  the  Time. 

Example.  —  Paid  $4.10  for  discounting  a  note  of  $150,  at 
8^.     For  what  time  was  it  discounted  ? 

SOLUTION.  Explanation.  —  The  int. 

$150  X  .08  =  $12,  Dis.  or  dis.  of  $150  for  1  yr.,  at 

$4.10  -r-  $12  =  ^glgpo.  or  ^  yr.        8%  =  $12.     Since  the  given 

tV^t  yr.  =  4  mo.  3  da.  ^^'  ^  ^^•^^'  *^^  *^«  ^^ 

^       "^  be  that  part  of  a  yr.  repre- 

sented   by  the    quotient  of 
$4.10  -4-  $12,  or  yVjT*  o^"  xVff  TT.  =  4  mo.  3  da..  Time. 

403.   KuLE. — Divide  the  given  discount  hy  the  discount 
of  the  face  at  the  given  rate  for  one  year  ;  the  quotient  is  the 


time,    ( 73 =  t. ) 

\B  X  r  I 


PBOBJLEMS. 

1.  The  face  of  a  note  is  $300 ;  discount,  $4.65  ;  rate  %,  6. 
How  long  has  it  to  run  ?  Ans.  93  da. 

2.  The  face  of  a  note  is  $500 ;  discount,  $2.83  ;  rate  %,  6. 
In  what  time  will  it  be  due  ?  Ans.  34  da. 

3.  The  face  of  a  note  is  $250 ;  discount,  $5.81 ;  rate  %,  9. 
How  long  before  it  matures  ?  Ans.  S  mo.  3  da, 

4.  The  face  of  a  note  is  $350 ;  discount,  $11.96  ;  rate  %,  10. 
How  long  has  it  to  run  ?  A7is.  4  mo.  3  da. 


354  PERCEKTAGE. 

6.  Its  face  is  $800 ;  discount,  814.47 ;  rate  ^,  7  ? 

A71S.  3  mo.  3  da. 

6.  Its  face  is  $1000 ;  discount,  $13.06 ;  rate  %,5? 

Arts.  94  da. 

7.  Its  face  is  $100 ;  discount,  $1.55 ;  rate  ^,  6  ? 

Ans.  3  mo.  3  da. 


OOrue  or  JEquitahle  Discotmt. 

True  Discount^  like  Banh  Discount,  has  Five  Cases, 
corresponding,  with  but  httle  variation,  to  the  Five  Cases  in 
Simple  Interest. 

The  attentive  pupil  -will  have  noticed  that  the  party  who  sells  his 
note  at  a  bank  pays  a  discount  not  only  on  the  mcmey  he  recei'oes  (the 
proceeds),  but  also  on  what  he  allows,  or  pays,  the  hank  for  the  v-se  of 
the  money.  If  the  discount  were  for  a  long  time,  the  injustice  of  this 
would  be  quite  evident. 

For  Example.— Suppose  a  man  wished  to  sell  his  note  of  $1000, 
running  10  years,  at  10^  discount.  How  much  would  the  note  on  this 
supposition  yield  him  ?  Ans.  Nothing.  For  at  10%  for  10  yr.  the  dis- 
count of  $1  =  $1,  and  the  discount  of  $1000  =  $1000,  and  $1000 
(Face)  -  $1000  (Discount)  =  0: 

Banks,  however,  do  not  discount  notes  for  long  periods. 

In  True  Discount,  discount  is  reckoned  only  on  the  »um  paid  for 
the  note,  mortgage,  or  other  obligation. . 

403.  Case  1.— Present  Worth,  Bate  %,  and 
Titne  given,  to  find  the  Discount,  or  the  Face* 

404.  EuLE. — Multiply  the  present  worth  ty  the  rate,  and 
faat  product  hy  the  time,    {B  x  r  x  t  =  P,)     (353.  I.) 

To  find  the  Fa4^,  add  to  the  present  worth,  the  discount. 

405.  Case  II. — Present  Worth,  Time,  and 
Discount  given^  to  find  tlie  Rate  %• 


DISCOUNT.  355 

406.  EuLES. — Take  such  a  part  of  100%  as  the  given 
discount  is  of  the  product  of  the  present  worth  and  time. 

Divide  the  given  discount  hy  the  interest  of  the  present 
toorth  for  the  given  time  at  1%  ;  the  quotient  is  the  rate  %. 

in — ^? — i  =  ^l     (355.) 
\B  X  1%  xt  J     ^         ' 

407.  Case  111.— Discount^  Hate  %,  and  Time 
given,  to  find  Present  Worth. 

408.  Rule. — Divide  the  given  discount  hy  the  interest 
of  SI  for  the  given  time  and  rate;  the  quotient  is  the  present 

worth.     [^^  =  b)     (357.  L) 

409.  Case  lY.—Face  or  Ainounty  Hate  %,  and 
Thne  given,  to  find  Present  Worth, 

410.  Eule. — Divide  the  face,  or  amount,  hy  the  amount 
of  $1  for  the  given  time  and  rate  ;  the  quotient  is  the  present 

worth. 


(jT^x7  =  ^')  (3S9-: 


In  importance,  this  Case  would  come  first ;  but  in  order  that  the 
Cases  of  Interest,  Bank  Discount,  and  True  Discount  may  be  uniform, 
it  has  been  thought  best  to  place  them  in  the  present  order. 

Of  course,  the  Present  Worth  having  been  found,  the  Discount  may 
be  obtained  by  subtracting  the  Present  Worth  from  the  Face. 

411.  Case  Y.  —  The  Present  Worth,  Hate  %, 
and  Discount  given,  to  find  the  Time. 

412.  Rule. — Divide  the  given  discount  by  the  interest 
of  the  Present  Worth,  at  the  given  rate,  for  1  ye^r  ;  the  quo- 
tient is  the  time,     i ^  =  t.\    (361.) 


356  PERCENTAGE. 

PR  O  It  T.EMS. 

Case  I. — Find  the  discount  and  face  of  a  note  whose 

1.  Present  Worth  is  $150 ;  rate  %,  6 ;  time,  4  mo. 

Ans.  $3;  $153. 

2.  Present  Worth  is  $1200 ;  rate  %,  7  ;  time,  6  mo. 

Ans.  $42;  $1242. 

3.  Present  Worth  is  $600  ;  rate  %,  4 ;  time,  1  mo. 

Ans.  $2;  $602. 

4.  Present  Worth  is  $3802.281;  rate  %,  7;  time,  4  yr. 
6  mo.  Ans.  $1197.719  ;  $5000. 

Case  II. — Find  the  rate  %  discount  of  a  note  whose 

5.  Present  Worth  is  $400  ;  discount,  $5 ;  time,  3  mo. 

Ans.  6%. 

6.  Present  Worth  is  $325 ;  discount,  $6.50 ;  time,  2  mo. 
12  da.  Ans.  10%. 

7.  Present  Worth,  £800;  discount,  £12  8s.;  time,  3  mo. 

3  da.  Ans.  Q%- 

8.  Present  Worth,  $3802.281 ;  discount,  $1197.719 ;  time, 

4  yr.  6  mo.  A7is.  '7%. 
Case  III. — Find  the  present  worth  of  a  note  whose 

9.  Discount  is  $3;  time,  4  mo.;  rate  %,  6.     Ans.  $150. 

10.  Discount  is  $2.60 ;  time,  4  mo.  10  da. ;    rate  %,  9. 

Ans.  $80. 

11.  Discount  is  $452.85  ;  time,  3  yr.  9  mo. ;  rate  %,  7. 

Ans.  $1725.15. 

12.  Discount  is  $91.07  ;  time,  2  yr.  8  mo. ;  rate  %,  6. 

Ans.  $569.19. 
Case  IV. — Find  the  present  worth  of  a  note  whose 

13.  Face  is  $602  ;  time,  1  mo. ;  rate  %,  4.      Ans.  $600. 

14.  Face  is  $2178;  time,  3  yr.  9  mo. ;  rate  %,  7. 

Ans,  $1725.15. 


DISCOUNT. 


357 


15.  Face  of  a  note  is  $5000 ;  time,  4  yr.  6  mo. ;  rate  %,  7. 
What  is  the  present  worth  ?  Ans,  $3802.281. 

16.  Face  of  a  note  is  $5560;  time,  1  yr.  6  mo.;  rate  %,  7. 
Find  the  present  worth.  Ans.  $5031.67. 

Case  V.— To  find  time. 

17.  Present  Worth  of  a  note  is  $1200;   discount,  $42; 
rate  %,  7.    What  is  the  time  ?  Ans.Q  ma 

18.  Present  Worth  of  a  note  is  $820;  discount,  $12.71; 
rate  %,  6.     How  long  does  it  run  ?  Ans.  3  mo.  3  da. 

19.  Present  Worth   of  a  note  is  $3802.281;    discount, 
$1197.719;  rate  %,  7.    Find  fche  time.       Ans.  4  yr.  6  ma 

20.  Present  Worth  of  a  note  is  $3000 ;  discount,  $540 ; 
rate  %,  5.    What  is  the  time  ?  Ans.  Z  yr.  7  mo.  6  da. 


SECTION    VIII. 


■ST'OCKS  i 


^ 


413.  StocJcs  are  the  funds,  or  capital,  of  Corporations. 

414.  A  Corporatiofi  is  a  company  authorized  hy  a 
charter,  or  law,  to  act  and  be  considered  as  a  single  indi- 
vidual, 

415.  A  Charter  is  a  legal  instrument  incorporating 
certain  persons,  and  defining  the  powers  and  duties  of  the 
corporation. 

416.  Stocks  are  usually  represented  by  a  certain  num^ 
ber  of  equal  parts  called  Shares,  which  are  usually  $100 
each,  though  they  may  be  of  any  value  agreed  upon. 


358  PEKCENTAGE. 

417.  Certificates  of  Stocky  called  Scrij)^  are  state- 
ments showing  the  number  of  shares  which  an  individual 
owns. 

418.  Stockholders  are  the  owners  of  the  shares  of 
stock. 

419.  Stocks  are  at  par  when  the  shares  sell  for  their 
face. 

420.  Stocks  are  above  par,  or  at  a  premiiun, 

when  the  shares  sell  for  more  than  their  face. 

431.  Stocks  are  heloiv  pwr,   or  at  a  discount, 

when  the  shares  sell  for  less  than  their  face. 

422.  The  Market  Value  of  stocks  is  the  price  at 
which  they  sell. 

423.  Dividends  are  portions  of  the  earnings  divided 
among  the  stockholders  in  proportion  to  the  stock  they  own. 

424.  Assessments  are  contributions  made  by  stock- 
holders to  supply  the  deficiency,  when  from  any  cause  the 
expenses  are  in  excess  of  the  income. 

425.  I^onds  are  written  obligations  by  which  parties 
bind  themselves  to  pay  specified  sums,  at  or  before  certain 
times  specified  in  the  bonds. 

Bonds  are  of  various  kinds :  United  States  Bonds,  State, 
County,  City,  Borough,  Eailroad,  Express,  Bridge,  Mining 
Co.  Bonds,  &c. 

Of  these,  the  XJ.  S.  Bonds  are  probably  the  most  impor- 
tant.    The  principal  ones  are 

1.  "  6's  of  '81 ,"  bearing  6%  interest,  and  payable  in  1881. 

2.  "5-20's,"  bearing  6%  interest,  and  payable  in  not  less  than  5 
nor  more  than  20  yr.  from  their  date,  at  the  pleasure  of  the  govern- 
ment.    Interest  paid  semi-annually  in  gold. 


STOCKS.  359 

3.  "10-40's,"  bearing  5%  interest,  redeemable  after  10  yr.  from 
their  date  ;  interest  semi-annually  in  gold. 

4.  "5's  of  '81,"  bearing  ofo  interest,  redeemable  after  1881 ;  inter- 
est paid  quarterly  in  gold. 

5.  "l|'s  of  '86,"  bearing  41%  interest,  redeemable  after  1886; 
interest  paid  quarterly  in  gold. 

6.  "4's  of  1901,"  bearing  4%  interest;  the  principal  payable  after 
1901 ;  the  interest  paid  quarterly  in  gold. 

436.  In  Coni2)utati07is  of  Stocks  and  Bonds  it  is 
important  to  remember 

1st.  That  premium  and  discount  are  reckoned  as  a  percentage  of 
the  par  value,  and  not  the  market  value. 

2d.  That  Par  Value  is  the  same  as  Base  lq  Percentage. 

That  Rate  Per  Cent  is  the  same  as  llate  Per  Cent  in  Percentage. 

That  Premium,  or  Discount,  is  the  same  as  Percentage  in  Per- 
centage. 

That  Market  Value  is  the  same  as  Sum  or  Difference,  in  Per- 
centage. 

Hence,  in  Stocks  and  Bonds,  we  have  simply  to  apply  the  rules  in 
Percentage. 

PRO  B  L  EMS. 

1.  What  cost  30  shares  bank  stock,  par  value  $100,  at  \^% 
premium,  brokerage  ^%  ? 

Solution.— |100  x  30  x  1.09  +  (3000  x  .00^,  brokerage)  = 
$3277.50,  Ans. 

2.  What  cost  25  shares  railroad  stock,  par  value  150,  at 
12|^^  discount  ? 

Solution.— $50  x  25  x  .87|  =  $1093.75,  Ans. 

3.  What  cost  100  shares  P.  R.  R.  stock,  par  value  $50,  at 
9^%  premium;  brokerage,  ^%  ?  ^ns.  $5500.25. 

4.  What  cost  25  6%  U.  S.  Bonds,  par  value  $100,  at 
116|  ?  Ans.  $2903.13. 

5.  What  cost  30  $500  10-40's,  at  16|^  premium,  broker- 
age!^? Ans.  $17550. 


360  PEECEITTAGE. 

6.  What  cost  50  IIOOO  U.  S.  6's  of  1881,  at  21 J^  pre- 
mium  ?  Ans. 

7.  What  cost  30  shares  Michigan  Southern  R  R,  par 
value  $50,  at  11%  premium,  brokerage  1^^? 

Ans.  11681.88. 


8.  Invested  $12000  in  stock  whose  par  value  was  $10000. 
What  %  premium  did  I  pay  ? 

Solution.— $13000  —  $10000  =  $2000.  J^^ x  100%  =  20%, Ans. 

$10000 

9.  Paid  $800  for  20  shares  R  R  stock,  par  value  $50. 
What  was  the  rate  %  discount  ? 

Solution.— $50  x20=$1000.  $1000— $800= $200.   J^  x  100% 

=  30%,  ^715. 

10.  Paid  a  premium  of  $100  on  20  shares  of  mining  stock 
whose  par  value  was  $50  per  share.    What  was  the  rate  %  ? 

Ans.  10%. 

11.  Having  purchased  25  shares  of  a  quicksilver-mine,  at 
an  advance  of  $20  per  share  for  $1750,  what  %,  premium  did 
I  pay  ?  Ans.  A0%. 

12.  How  many  $100  10-40's,  at  a  premium  of  16^,  bro- 
kerage i%,  can  I  purchase  for  $4998. 75  ? 

Solution.— At  a  premium  of  16%,  a  $100  bond  is  worth  $116;  to 
this  add  \%,ot  $.35  for  brokerage,  and  the  entire  cost  is  $116.25. 
For  $4§98.75,  I  can  purchase  as  many  $100  bonds  as  $116.35  is  con- 
tained times  in  $4998.75,  which  are  43  times.  Therefore,  43  is  the 
number  of  bonds. 

13.  How  many  dollars  in  gold,  at  a  premium  of  12^^, 
brokerage  ^^,  can  be  bought  for  $1130  in  greenbacks? 

Ans.  $1000. 


STOCKS.  361 

14.  How  many  dollars  in  silver,  at  a  discount  of  2^,  bro- 
kerage ifc,  can  be  purchased  for  19812.50  ?    Ans.  $10000. 

15.  How  many  shares  P.  K.  E.  stock,  par  value  $50,  can 
I  purchase  for  $6875,  when  they  are  at  a  premium  of  9|-^, 
and  brokerage  costs  me  i%  ?  A7is.  125  shares. 

16.  How  many  $500  bonds  of  '81,  at  a  premium  of  16-^^, 
brokerage  1^,  can  be  purchased  for  $58250  ? 

A71S.  100  bonds. 

17.  At  what  price  will  a  6%  bond  of  $500  yield  8%  ? 

Solution. — 6%  of  $500  =  $30.  At  8%  it  will  require  as  many 
dollars  to  yield  $30,  as  $.08  is  contained  times  in  $30,  that  is,  375 
times ;  or,  at  $375,  a  6%  $500  bond  will  yield  Sfo. 

At  what  price  will  a 

18.  10^  bank  stock,  par  value  $100,  yield  S%  interest  ? 

Ans.  $125. 

19.  S%  raiboad  bond  of  $1000,  yield  6%  ? 

Ans,  $1333J. 

20.  10-40  U.  S.  $500  bond,  yield  6^%,  when  gold  is  at  a 
premium  of  12 J^^  ?  Ans.  $432.69. 

21.  6%  mining  stock,  par  value  $50,  pay  S%  ? 

Ans.  $31.25. 

22.  Which  is  the  better  investment,  5-20*s  at  122,  or 
10-40's  at  116  ?  Ans.  The  former. 

23.  Which  is  the  better  investment,  "  4|'s  of  '86  "  at  par, 
or"4'sofl901"atlH^dis.  ? 

24.  Which  is  the  better  investment  'aO-40's"  at  112,  or 
"  4i's  "  at  par  ?  Ans,  The  latter. 


SECTION    IX 


jj 


(T^^S)  ®  (f)  ^i:^^^ 


]BX.C'H^M<e:iE    ll 


L 


•^j^ 


437.  Exchange  is  a  method  of  making  payment  at  a 
distant  place  by  means  of  drafts^  or  hills  of  exchange. 

428.  A  Draft,  or  Sill  of  ExcTianffe,  is  a  written 
order,  directed  by  one  party  to  another,  to  pay  money  to  a 
third  party. 

The  Drawer  is  tlie  maker  of  the  order ;  the  JDratvee  is  the 
party  to  whom  the  order  is  directed ;  the  Payee  is  the  party  to 
whom  payment  is  ordered ;  the  Buyer,  or  Remitter,  is  the  party 
who  sends  the  draft. 

429.  An  Inland  Draft,  or  Dill  of  Exchange, 

is  one,  of  which  the  drawer  and  drawee  reside  in  the  same 
country.  This  is  sometimes  called  home,  or  domestic  ex- 
change. 

430.  A  Foreign  Dill  of  Exchange  is  one,  of  which 
the  drawer  and  drawee  reside  in  different  countries. 

Exchange  is  at  par  when  the  price  of  a  draft  is  the  face  of  it ;  at  a 
premium,  or  above  par,  when  the  price  of  it  is  more  than  its  face ;  and 
at  a  discount f  or  below  pa/r,  when  the  price  of  it  is  less  than  its  face. 

431.  The  Date  of  Exchange  is  a  percentage  of  the 
face  of  the  draft,  reckoned  as  premium  or  discount. 

432.  The  Course  of  Exchange  is  1,  increased  by 
the  rate  of  premium,  or  diminished  by  the  rate  of  dis- 
count. 


EXCHAJSTGE.  363 

That  place,  or  region,  which  buys  of  another  more  than  it  sells  to 
it,  pays  a  premium  for  drafts  upon  it,  because  the  indebted  place  has 
not  by  its  business  created  credits  enough  in  the  other  place  to  bal- 
ance its  debts,  and  must  send  money  there  at  some  cost. 

That  place,  or  region,  which  sells  to  another  more  than  it  buys 
from  it,  gets  drafts  upon  it  at  a  discount,  because  the  former  place 
has  more  drafts  upon  the  latter  than  it  needs  to  balance  its  own 
debts,  and  the  excess  must  be  collected  at  some  cost. 

If  a  bill  of  exchange  is  paid  when  due,  it  is  said  to  be  honored  ;  if 
not,  it  is  said  to  be  dishonored. 

A  bill  is  accepted  when  the  drawee  agrees  to  pay  it  at  maturity. 
He  usually  does  this  by  his  signature  to  the  word  "  Accepted,"  writ- 
ten across  the  face  of  the  bill. 

433.  A  Sight  Draft  is  one  which  is  to  be  paid  when 
presented. 

434.  A  TlTYie  Draft  is  one  payable  at  some  future 
time  named  in  the  draft. 

435.  Three  days  of  grace  are  allowed  on  Time 
Drafts,  but  not  on  Sigid  Drafts. 

In  computing  the  value  of  drafts  or  bills  of  exchange,  allowance 
must  be  made  for  the  length  of  time  elapsing  from  the  time  the  draft 
is  made  until  it  is  paid.  For  example  :  on  account  of  the  distance  of 
the  drawee,  or  because  the  buyer  may  not  wish  to  draw  the  money,  a 
month  or  more  may  elapse  before  the  draft  is  paid.  Under  such  cir- 
cumstances, when  known  at  the  time  the  draft  is  purchased,  the  buyer 
is  allowed  interest  for  the  intervening  time. 

The  Principles  and  Eules  of  Percentage,  with  which  the 
pupil  who  has  progressed  thus  far,  should  now  be  perfectly 
familiar,  are  applicable  to  calculations  in  Exchange,  the 
only  caution  needed,  perhaps,  being 

436.  That  all  calculations  are  based  upon  the  face  of 
the  draft'. 


364  PERCENTAGE. 

FM  O  B  LEM  S, 

1.  What  costs  a  sight  draft  on  New  Orleans  for  $5672,  at 
1^%  premium  ? 

Solution.— ($1  +  $.01^)  x  5672  =  $5742.90.  Ana. 

2.  What  costs  a  sight  draft  for  $7560,  on  Chicago,  at  i% 
discount  ? 

Solution.— ($1  -  $.00^)  x  7560  =  $7550.55.  Ans. 

What  costs  a  sight  draft 

3.  For  1500,  at  1^%  premium?  Ans.  $507.50. 

4.  For  $7000,  at  i%  discount  ?  Ans.  $6965. 

5.  For  $17500,  at  J^  premium  ?  Ans.  $17631.25. 

6.  For  $6572,  at  1^%  discount  ?  Ans.  $6489.85. 

7.  What  costs  a  draft  on  Chicago  for  $3340,  drawn  "30 
days  after  sight,"  premium  ^^c,  interest  allowed  at  8%  ? 

Solution.— ($1  +  $.00|  —  $.00ii)  x  3340  =  $334473.  Ans. 

Observe  that  the  premium  and  interest  are  calculated  on  the  face 
of  the  draft. 

8.  What  cost  a  draft  payable  after  sight,  60  da.  for  $1000, 
at  i%  prem.,  interest  9%  ?  Ans.  $989.25. 

9.  90  da.  for  $7000,  at  f  ^  prem.,  interest  7%  ? 

A71S.  $6925.92. 

10.  30  da.,  for  $7600,  at  1%  dis.,  interest  6%  ? 

Ans.  $7482.20. 

11.  90  da.,  for  $1700,  at  1^%  dis.,  interest  12^  ? 

Ans.  $1621.80. 

12.  What  costs  a  sight  draft  on  London  for  £200  ? 
Solution.— $4.8665  x  200  =  $973.30,  Ans.    (See  Art,  195.) 

13.  What  costs  a  sight  bill  on  Liverpool  for  £650  ? 

Ans.  $3163.23. 


EXCHAKGE.  365 

14.  What  costs  a  60  da.  bill  on  London  for  £7000,  interest 
at  6^  ? 

Solution.— $48665  x  (1  -  .0105)  x  7000  =  $33707.81,  Ans. 

15.  What  costs  a  30  da.  bill  on  Glasgow  for  £500,  interest 
at  10^  ?  Ans.  $2410.95. 

16.  What  costs  a  60  da.  bill  on  Paris  for  5000  fr.,  interest 
at  12^  ? 

Solution.— $.193  x  (1  -  .021)  x  5000  =  $944.74,  Am.    (See  Ai^t. 
197. 

17.  What  costs  a  90  da.  bill  on  Havre  for  1000  francs, 
interest  at  9%  ?  Ans.  $188.51, 

What  is  the  face  of  a  sight  draft  which 

18.  At  1^%  prem.,  costs  $507.50  ? 
Solution.— $507.50  -4-  (1  +  .Oli)  =  $500,  Ans. 

19.  At  1%  dis.,  costs  $6965  ?       ,  Ans.  $7000. 

What  is  the  face  of  a  time  draft  which 

20.  At  ^%  prem.,  interest  6%  for  60  da.,  costs  $994.50  ? 
Solution.— $994.50  h-  ($1  -  $.0105  +  $.005)  =  $1000,  Ans. 

21.  At  1%  dis.,  int.  at  6%  for  30  da.,  costs  $7482.20  ? 

A71S.  $7600. 

22.  At  11%  dis.,  int.  at  12^  for  90  da.,  costs  $1621.80  ? 

Ans.  $1700. 

23.  A  sight  bill  on  Liverpool  cost  $3163.225.    What  was 
its  face  ? 

Solution.— $3163.225  -f-  4.8665  =  £650,  Ans. 

24.  A  sight  bill  on  Glasgow  cost  $973.30.    What  was  the 
face  of  the  bill  ?  Ans.  £200. 

25.  A  60  da.  bill  on  Liverpool,  at  6%  int.,  cost  $33707.81. 
What  was  the  face  of  the  bill  ? 


Solution.— $33707.81  -^  (1  -  .0105)  x  $4  8665  =  £7000,  Ans. 


PERCEI^TAGE, 


26.  What  is  the  face  of  a  draft  on  Paris  for  60  da.,  at  12^, 
costing  1944.735? 

Solution.— $944,735  -5-  (1  -  .031  )  x  $.193  =  5000  fr.,  Ans 

27.  What  is  the  face  of  a  90  da.  bill  on  Havre,  int.  at  9%, 
costing  $188,513  ?  A?is.  1000  fr. 

The  following  Table  exhibits  the  value  of  the  coins  of  a 
number  of  foreign  countries,  in  U.  S.  money,  as  proclaimed 
by  the  Secretary  of  the  Treasury,  Jan.  1,  1878 : 


Florin,  Austria   $.453 

Franc,  Belgium 193 

Dollar,  Bolivia 965 

Milreis  of  1000  reis, 

Brazil .545 

Dollar,  British  North 

America $1.00 

Peso,  Bogota 965 

Dollar,  Cent.  America,   .918 

Peso,  Chih 912 

Crown,  Denmark 268 

Dollar,  Ecuador 918 

Pound  of  100  piasters, 

Egypt $4,974 

Franc,  France 193 

Pound  Sterling,  Great 

Britain $4.866| 

Drachma,  Greece 193 

Mark,  German  Empire,   .238 


Yen,  Japan $.997" 

Eupee   of   16    annas, 

India .436 

Lira,  Italy $.193 

Dollar,  Liberia $1.00 

Dollar,  Mexico 998 

Florin,  Netherlands. .     .385 

Crown,  Norway 268 

Dollar,  Peru 918 

Milreis,  Portugal $1.08 

Eouble,  Russia 734 

Dollar,  Sandwich  Is. .  $1.00 

Peseta,  Spain 193 

Crown,  Sweden 268 

Franc,  Switzerland. . .     .193 

Mahbub,  Tripoli 829 

Piaster,  Tunis 118 

Piaster,  Turkey 043 

Peso,  TJ.  S.  Columbia,    .918 


28.  What  is  the  face  of  a  60  da.  draft  on  Lisbon,  Portugal, 
int.  at  10^,  costing  $2652.75  ?  Ans.  2500  milreis. 


m  WMXMm  MMW  BUT'IE m 


T 

437.  A  Tax  is  a  sum  of  money  required  from  individ- 
uals by  Government  for  public  purposes. 

Taxes  are  either  Direct  or  Indirect 

438.  A  Direct  Tax  is  a  tax  assessed  directly  upon 
the  property  or  persons  of  taxable  individuals. 

439.  An  Indirect  Tax  is  a  tax  on  articles  of  con- 
sumption in  their  transit  from  one  person  to  another. 

Direct  taxes  are  either  Poll  or  Property-Taxes. 

440.  A  Poll' Tax f  or  Capitation- Tax^  is  a  tax 

imposed  equally  on  taxable  persons,  without  regard  to  their 
property. 

A  PoUy  in  law,  is  a  taxable  person. 

441.  A  Property- Tax  is  a  tax  assessed  at  a  given 
rate  on  the  value  of  property. 

Property  is  either  Beat  Estate  or  Personal. 

443.  Real  JEstate  is  fixed  property,  such  as  lands, 
houses,  &c. 

443.  Personal  Property  includes  all  kinds  of 
property  not  fixed,  such  as  furniture,  vehicles,  live-stock, 
goods,  &c. 

An  Inventory  is  a  list  of  articles. 

367 


368  PERCENTAGE. 

444.  Direct  Taxes. 

To  assess  a  direct  jjroperty  tax. 
Example  1.— A  town  is  to  raise  15743.64 +  .     The  tax- 
ables  are  831,  their  property  is  assessed  as  worth  11637545, 
and  the  poll-tax  is  $1.     What  is  the  rate  of  property-tax  ? 

Explanation.— Since  each 

SOLUTION.  poll  pays  $1,  831    polls  pay 

831  X  II  =  $831,  Poll-tax.  $831,  leaving  $4912.64  to  be 

$5743.64  -  1831  =  $4912.64       ^^^^  ^rZ^JX 
$4912.64  -^  $1637545  =  .003        pays  $.008.  that  is,  the  tax  is 

8  mills  on  the  dollar. 

445.  EuLE. — From  the  sutn  to  he  raised  subtract  the 
amount  to  he  raised  hy  poll-tax.  The  remainder  divided  hy 
the  assessed  value  of  taxahle  property,  will  give  the  rate  on 
one  dollar. 

Multiply  the  tax  on  one  dollar  hy  the  assessed  value  of  a 
person^ s  property,  and  to  the  product  add  his  poll-tax,  if  any. 

PROBLEMS. 

1.  Taxables,  1224 ;  poll-tax,  75^ ;  amount  to  be  raised, 
$3418 ;  property  assessed  at  $500000.  What  is  the  tax  on 
$1  ?  Ans.  5  mills. 

2.  What  is  A's  tax,  whose  property  is  assessed  at  $5725, 
and  who  pays  for  2  polls  ?  A7is.  $30.13. 

3.  What  is  B's  tax,  whose  property  is  assessed  at  $2380, 
and  who  pays  for  1  poll  ?  Ans.  $12.65. 

4.  What  is  C's  tax,  whose  property  is  assessed  at  $9775, 
and  who  pays  for  3  polls  ?  Ans.  ^51.13. 

5.  A  town  containing  453  taxables  and  whose  property 
real  and  personal  was  assessed  at  $2560000,  was  compelled 
to  raise  for  expenses  in  1876,  $11795.20.     Making  an  allow- 


TAXES     AND     DUTIES.  369 

ance  of  6%  for  collecting,  and  d%  for  lost  taxes,  what  must 
the  assessment  be  ?  Ans,  ^%,  or  $12800. 

6.  What  will  be  A's  tax,  whose  personal  property  is  assessed 
at  $3150,  and  real  estate  at  $7560?  Ans,  $53.55. 

Duties,  or  Indirect  Taxes, 

446.  Duties  are  taxes  imposed  upon  merchandise  in 
transportation. 

447.  Customs  are  duties  on  goods  imported  or  ex- 
ported. 

448.  Excise  is  an  inland  duty  on  articles  manu- 
factured. 

449.  A  Tariff  is  a  list  of  duties. 

450.  Hevenue  is  income  derived  from  duties  and 
other  sources. 

451.  A  Custom  Mouse  is  a  house,  or  office,  where 
the  Goyernment  business  concerning  imports,  exports,  and 
duties,  is  performed. 

453.  An  Invoice^  or  a  Manifest,  is  a  written  ac- 
count of  the  articles  of  merchandise  transported. 
Duties  are  either  Specific  or  Ad  Valorem. 

453.  A  Specific  Duty  is  a  duty  on  a  definite  quan- 
tity of  an  article,  without  reference  to  its  value. 

454.  An  Ad  Valorem  Duty  is  a  duty  imposed  at  a 
certain  rate  per  cent  of  the  cost  of  an  article. 

In  estimating  specific  duties,  certain  deductions,  op  allowances,  are 

sometimes  made  for  packages  and  waste. 

455.  HreaUage  is  a  deduction  for  loss  by  breaking. 

456.  Leakage  is  a  deduction  for  loss  by  leaking. 


370  PERCEKTAGE. 

457.  Tare  is  an  allowance  for  the  weight  of  the  thing 
containing  the  goods. 

458.  I>rafty  or  Tret^  is  an  allowance  for  waste  in 
handling. 

459.  Gross  Weight  is  the  weight  of  the  goods  and 
the  thing  which  contains  them. 

460.  Wet  Weight  is  weight  remaining  after  the  de- 
ductions. 

In  the  collection  of  specific  duties,  after  all  allowances  are  made, 
the  problem  reduces  to  one  of  simple  multiplication. 

In  calculating  ad  valorem  duties,  after  the  deductions  are  made  the 
problem  becomes  one  in  percentage. 

Pit  OBLEMS, 

1.  What  is  the  duty  on  150  boxes  raisins,  16  lb.  per  box, 
at  %f  per  lb. ;  tare  25^  ? 

Solution.— 150  boxes  of  16  lb.  each  —  2400  lb.  The  tare,  25%  of 
2400  lb.  =  600  lb.  Subtracting  the  tare  from  2400  lb  ,  there  remain 
2400  lb.  —  600  lb.  =:  1800  lb.,  on  which  to  collect  the  duty  of  2^ 
per  lb.    2i^  x  1800  =  3600)*  =  $36,  Arts. 

2.  What  is  the  duty  on  250  T.  steel,  at  $45  per  T.? 

Ans.  $11250. 

3.  What  is  the  duty  on  50  bbl.  sugar  of  250  lb.  each,  tare 
10^,  atlj^?  Jt^s.  $168.75. 

4.  What  is  the  duty  on  carpeting  invoiced  at  £560  5s.,  at 


Solution.— £560  5s.  =  £560.25.    $4.8665  x  560.25  =  $2-726.4566}, 
which  at  35%  gives  $954.26,  Ans. 

5.  What  is  the  duty  on  50  crates  of  crockery,  Invoiced  at 
$9500 ;  breakage,  10^ ;  at  30^  ad  valorem  ?    Ans.  $2565. 


TAXES     AND      DUTIES 


371 


6.  What  is  the  duty  on  400  doz.  bottles  of  Porter,  in- 
voiced at  $350 ;  breakage,  10^ ;  at  25^  ad  valorem  ? 

Ans.  178.75. 

7.  What  is  the  duty  on  leather  invoiced  at  $9740 ;  dam- 
ages, 40^;  at  15^  ad  valorem  ?  Ans.  $876.60. 

8.  A  manufacturer  paid  to  the  U.  S.  Government  $2500 
duties;  excise  taxes,  $3500;  income  tax,  $450;  and  used 
revenue  stamps  to  the  value  of  $1750,  and  postage-stamps 
worth  $963.  How  much  did  he  contribute  to  the  support 
of  the  National  Government  ?  Ans.  $9163. 


SECTION    XI 


^ 


WAMTmmmmmiW'  ■  ■|! 


io- 


461.  Partnership  is  the  association  of  two  or  more 
persons  who  unite  their  money  and  labor  in  the  transaction 
of  business. 

Partnersliip  is  of  two  kinds.  Simple  and  Compound. 

Simple  Partnersliip  is  that  in  which  the  capital  of  each  partner  is 
employed  for  the  same  period  of  time. 

Compound  Partnership  is  that  in  which  the  capital  of  each  partner 
is  employed  for  uneqvM  periods  of  time. 

462.  Partners  are  persons  associated  in  business. 

463.  A  business  association  is  called  a  Company ^ 
Firnif  or  Souse. 

464.  Capital,  or  Stock,  is  the  property  employed  in 

business. 


372  PERCENTAGE. 

465.  A  Dividend  is  that  which  is  divided  among 
partners. 

466.  An  Assessment  is  a  demand  for  money  to  be 
paid  by  each  partner,  for  his  share  of  the  expenses,  or 
losses. 

46T.  The  Liabilities  of  a  company  are  its  debts. 

PJt  O  BL  EMS, 

1.  A  and  B  enter  into  partnership.  xVs  capital  is  $5000; 
B's,  $3000.  They  gain  $2400.  What  is  each  partner's 
share  ? 

Explanation. — We 

SOLUTION.  find  the  amount  of  cap- 

A's  capital  =  $5000  ital  invested   by  both 

B's         "        =  $3000  partners  to  be 


— the  entire  gain  is  5^2400. 

A  and  B's  capital  =  $8000  we  find  the  rate  % 

A  and  B's  gain      =  $2400  gain  by  Case  II.  in 

M«  X  100^  =  30^  gain  (399).  Xar^iX 

$5000  X  .30  =  $1500,  A's  gain  (29')').  Case  I.  of  Percentage. 

$3000  X  .30  =       900,  B's      "  ^^^^^  ^1^°  Examples  7, 

^  8,  and  9,  page  286.) 

2.  Two  partners,  M  and  N,  having  gained  $2700,  divided 
it  with  reference  to  the  capital  employed  by  each.  How 
much  will  they  each  receive,  provided  M's  capital  is  $9000 
and  N's  $18000  ?  Aiis.  M,  $900,  and  IST,  $1800. 

3.  Messrs.  Jones,  Smith  &  Simpson  lost  in  business  $7500. 
What  was  the  loss  of  each,  on  the  supposition  that  Jones' 
capital  was  $15000,  Smith's,  $20000,  and  Simpson's, 
$25000  ? 

^wg.  Jones,  $1875  ;  Smith,  $2500  ;  Simpson,  $3125. 


PARTKERSHIP.  373 

4.  Brown  and  Long  shared  £7284  12s.  profits  of  specula- 
tion. B's  stock  was  £5525,  L's  £7735.  Find  the  share  of 
each.  Ans.  B,  £3035  5s.;  L,  £4249  7s. 

5.  The  firm  of  Dayton  &  Co.  dissolved,  dividing  £7896 
18s.  6d.  D's  stock  was  £4000,  and  each  of  his  two  partners 
had  £1000.     Find  the  share  of  each. 

Ans.  D,  £5264  12s.  4d.;  the  others  each,  £1316  3s.  Id. 

6.  In  a  partnership  between  A,  B  and  0,  A  invested 
$3000  for  6  mo.;  B,  $5000  for  8  mo.;  and  C,  $6000  for 
7  mo.      They  gained    $5000.     What  was  each   partner's 

share  ? 

Explanation.  —  $3000 

SOLUTION.  employed  for  6  mo.  =  6  x 

$3000  X  6  =  $18000,  A'S  for  1  mo.     $3000,  or  $18000  employed 

$5000  X  8  r=  $40000,  B's    "     "  for  1  mo. ;  $5000  for  8  mo. 

$6000  X  7  =  $42000,  C's    "    "  =  ^  ^  ^^^^^'  ^'  ^^^^  ^^^ 
1  mo. ;  and  $6000  for  7  mo. 

$100000.  =  $42000  for  12mo.  Hence, 

,,  the  combined  capital  of  A, 

/^'  B,  and  C  was  equivalent 

$18000  X  .05  =     $900,  A's  gain,  to  $100000  for  1  mo.    On 

$40000  X  .05  =  $2000,  B's   "    H'  ^^T!^  ^^l^   r"" ' 

^  '  $5000,  or  5%.     Therefore, 

$42000   X  .05  =  $2100,  C's       "  each  gained  5%  on  what 

was  equivalent  to  his  stock 
for  1  mo. ;  and  A  receives  5%  of  6  x  $3000  ;  B,  5%  of  8  x  §5000;  and 
C,  5%  of  7  X  $6000. 

7.  Taft  and  Rodgers  divided  $15000  July  1,  1875.  T  be- 
gan Oct.  1,  1874,  with  $24000,  and  took  in  R,  Feb.  1,  1875, 
with  $16800.     Find  their  shares. 

A71S.  T,  $10800 ;  R,  $4200. 

8.  M,  N,  and  P  hired  a  pasture  for  $68.40.  M  pastured 
1  cow  30  weeks;  N,  1  cow  30  weeks,  and  another  20  weeks; 
and  P,  2  cows  30  weeks,  and  1  cow  12  weeks.  What  should 
eacli  pay  ?  Ans.  M,  $13.50 ;  N,  $22.50  ;  P,  $32  40. 


t'oUU  X  100^ 


374  PERCENTAGE. 

9.  Siemon  and  Bright  began  business  Apr.  1,  1873,  each 
with  a  capital  of  $10000.  They  took  in  Williams  Jan.  1, 
1874,  with  $5000.  They  dissolved  partnership  July  1,  1875, 
sharing  $31500  profit.     Find  the  share  of  each  ? 

Ans.  S  and  B,  each  $13500 ;  W,  $4500. 

10.  Macnum  &  Seward  began  business  Dec.  1,  1873 ;  M, 
with  $7500 ;  and  Seward,  with  $6000.  July  1, 1874,  M  took 
out  $1500.  They  dissolved  partnership  June  1, 1875,  paying 
$5662.50  debt.     What  had  each  to  pay  ? 

Ans.  M,  $2962.50;  S,  $2700. 

11.  A  school  had  an  attendance  one  term  equal  to  that  of 
1  pupil  2300  days.  Mr.  Caldwell  had  sent  2  pupils  each  90 
days,  and  1  pupil  60  days.  How  much  of  $149.50,  the  cost 
of  the  school,  should  he  bear,  if  pay  was  according  to 
attendance?  Ans.  $15.60. 

12.  Cook  and  Little  formed  a  partnership  for  1  year.  C 
had  I  of  the  capital  and  L  | ;  they  agreed  to  pay  Shai-p  J  of 
the  profits  for  managing  the  business.  At  the  end  of  the 
year  they  divided  $12000.    What  was  the  share  of  each  ? 

Ans.  S,  $3000;  C,  $5400;  L,  $3600. 

13.  Sill,  Eea  and  Collins  shared  $30000  profit  on  the  fol- 
lowing conditions :  for  every  $4  of  S's  stock  there  were  $3 
of  R's  and  $2  of  C's,  and  for  every  3  mo.  of  S's  time  there 
were  2  mo.  of  R's  and  1  of  C's.  What  was  the  share  of 
each  ?  Ans.   S,  $18000  ;  R,  $9000;  C,  $3000. 

14.  At  the  end  of  a  year,  A,  B  and  C  divided  profits 
amounting  to  $7500,  A  receiving  $3600;  B,  $2400;  C, 
$1500.  A  had  been  engaged  in  the  partnership  9  mo. ;  B, 
8  mo.,  and  C,  3  mo.     What  was  the  ratio  of  their  capital  ? 

Ans.  As  4,  3  and  5. 


SECTION    XII. 


BJAMKR-CJfP'TCY 


'^^ 


(5^ 


4i'QS.  Ba^ihrtiptcy^  or  Insolvency ,  is  iDability  to 
pay  debts,  through  lack  of  property. 

•  A  person  unable  to  pay  his  debts  is  called  a  bankrupt^  or  an  insol- 
vent.   The  liabilities  of  a  bankrupt  are  his  debts. 

469.  Debtors  are  persons  who  owe. 

470.  Creditors  are  persons  to  whom  debts  are  owed. 

4'71.  Assets  are  such  portions  of  property  as  can  be 
appropriated  to  paying  debts. 

472.  N'et  JProceeds  is  the  value  remaining  after  all 
necessary  deductions  have  been  made. 

473.  An  Assignee  is  a  person  selected  to  take  charge 
of  the  assets  of  an  insolvent,  or  bankrupt,  and  apply  them 
to  the  payment  of  the  creditors. 

PB.  O  B  r.  EMS. 

1.  Samuel  Jackson  placed  in  the  hands  of  an  assignee  for 
the  benefit  of  his  creditors,  a  farm  which  sold  for  $5600;  a 
mill  property  which  brought  $17000;  bank  stock  worth 
18000 ;  insurance  stock  worth  $5000.  His  entire  liabilities 
were  $69600.  The  assignee's  charges  were  $10680,  and 
other  expenses  were  $1720.  What  %  of  his  indebtedness 
did  he  pay  ?  A?is.  33^%. 

2.  How  much  did  S.  Thompson  receive  whose  claim  was 
$5000?  Ans.  $1666|. 

375 


376 


P  E  R  C  E  I^- T  A  G  E , 


3.  How  niiich   did  Amos  Miller  get  whose  claim  was 
118000?  Ans.  16000. 

4.  How  much  did  James  Horner  obtain  whose  claim  was 
$600?  Ans.  $200. 

SECTION    XIII. 


^■¥EiKA(3.E  ®F  PiAYMEMTSJ 


474.  Average  of  JPayments  is  a  process  of  finding 
the  time  for  paying  in  a  single  payment  two  or  more  debts 
which  are  due  at  different  times,  so  that  neither  debtor  nor 
creditor  may  suffer  loss  of  interest. 

475.  The  Averaf/e,   or   Equated   Tinier   is  the 

time  determined  by  this  process. 

476.  A  Focal  Date  is  any  assumed  date  with  which 
other  dates  are  compared. 

477.  The  Term  of  Credit  is  the  time  elapsing  be- 
tween the  contracting  of  a  debt  and  its  maturity. 

478.  The  Average  Term  of  Credit  is  the  average 
time  between  the  contracting  of  several  debts  and  their 
maturities. 

Average  of  Payments  is  of  two  kinds,  Simple  and  Compound. 

479.  In  SlTTiple  Average,  all  the  items  belong  to 
one  side  of  an  account,  that  is,  all  are  debts,  or  all  are  credits. 

480.  In  Compound  Average,  there  are  items  be- 
longing to  both  sides  of  an  account. 

481.  A7i  Account  is  a  formal  statement  made  by  one 
party  of  his  commercial  transactions  with  another. 


AVERAGE     AND     PAYMEN^TS.  377 

482.  An  account  has  two  sides,  one  the  debtor  side, 
marked  Dr.^  showing  items  of  debt ;  the  other,  the  credit 
side,  marked  Cr.,  showing  the  items  of  credit. 

483.  The  JBalance  of  an  account  is  the  difference 
between  the  amounts  of  the  Dr.  and  Or.  sides. 

Simple  Average, 

Example. — A  owes  B  $300,  to  be  paid  in  4  mo.,  and  $500 
to  be  paid  in  6  mo.  When  could  1800  be  paid  by  A  to  B, 
so  as  to  cause  no  loss  to  either  party  ? 

SOLUTION.  Explanation.— In  this  Exam- 

1300  X  .02  =    $6.00.  P^^  ^®  regard  A  as  having  the 

$500  X  .03  =  $15.00.  r  «''  ^^V  '  ""  "r  ^''' 

for  6  mo.     Now,  our  question  is, 

for  how  many  months  would  the 
use  of  their  sum,  $800,  be  an  equiv- 
alent. 

At  6%  (we  may  use  any  rate 
% ),  the  interest  of  $300  for  4  mo. 
equals  $6;  and  the  interest  of  $500  at  the  same  rate  %  for  6  mo. 
equals  $15;  together,  $6  +  $15  =  $21.  The  interest  of  $800  ($300  + 
$500)  for  1  yr.  at  6  %  is  $48.  Therefore,  instead  of  the  use  of  $300 
for  4  mo.  and  $500  for  6  mo.,  the  use  of  $800  for  |i  yr,,or5^  mo., 
(361)  is  an  equivalent. 

All  problems  in  Average  of  Payments  may  be  solved  by  the  fore- 
going method  ;  but  when  Interest  tables  are  not  at  hand,  the  method 
generally  adopted  is  exhibited  in  the  following 

solution.  Explanation.— The  use  of  $300 

$300    X   4  =  $1200  ^'^^  ^  ^^^'  ^^  equivalent  to  the  use  of 

tfj^nn  ^  (K  —  *qonn  ^  ^  ^^^^'  ^'^  ^^^^^'  ^^''  ^  '^''' '  *^^ 


$21. 

$800  X 

.06  = 

=  $48.     (353.) 

«yr.  = 

=  H 

mo.     (361.) 

$800  )  $4200  to  the  use  of  6   x  $500,  or  $3000, 

7j  for  1  mo.  ;  and  the  sum  of  these, 

5^  mo.  ^^,^QQ  ^  ^gO^^  ^^  ^^2QQ^  .g  ^^^^ 

alent  to  the  use  of  $4200  for  1  mo.     But  we  wish  to  know  for  how 


378  PEKCENTAGE. 

many  months  we  can  use  $800,  so  that  its  use  shall  be  equivalent  to 
the  use  of  $4200  for  1  mo.,  which  is,  as  we  have  just  shown,  equivalent 
to  the  use  of  $300  for  4  mo.,  and  $500  for  6  mo.  This  we  find  by 
dividing  $4200  by  $800.  Our  quotient,  5^,  gives  the  number  of 
months.    Hence,  the 

484.   Rule. — Multiply  each  item  iy  its  time,  and  divide 
the  sum  of  the  products  by  the  sum  of  the  item^. 


PROBLEMS. 

1.  John  Simpson  bought  a  farm,  for  which  he  was  to  pay 
$500  cash  ;  $600  in  6  mo. ;  $700  in  1  yr.,  and  $900  in  15  mo. 
He  afterward  agreed  to  pay  it  all  at  one  time.  Find  the 
time.  Ans.  9|^  mo. 

2.  Bought  goods  to  the  amount  of  $5000,  of  which  $1000 
was  to  be  paid  in  1  mo.,  $1000  in  2  mo.,  $2000  in  3  mo.,  and 
the  balance  in  6  mo.  If  I  gave  a  note  at  once  for  the  whole 
amount,  how  long  should  the  note  run  ?  Ans.  3  mo. 

3.  Bought  of  Joseph  Moon  &  Co.  goods  as  follows :  Jan. 
3,  1876,  $300,  on  1  mo.  credit;  Jan.  4,  1876,  $400,  on  2  mo. 
credit ;  Jan.  7,  1876,  $500,  on  3  mo.  credit.  At  what  time 
will  $1200  (  =z  $300  -f  ^400  +  $500)  cancel  the  debts  ? 

SOLUTION.  Explanation.  —  We  first 

Focal  Date,  Feb.  3,  1876.  add  the  terms  of  credit  to  each 


Feb.  3/76,  $300  x    0  = 


date  to  find  when  each  item 

is  due. 

Mar.  4/76,  $400  X  30  =  12000  We  then  fix  upon  Feb.  3 

Apl.  7/76,.  $500  X  64  =  32000         as  the  focal  date  (any  other 

~7ZZZ  ^  TTITT        date  would  answer  our  pur- 

1300  )  44000       ^^^  ^„a   fi^a  j,„^  J^„y 

37         days  intervene  between  Feb. 

Feb.  3  +  37  =  Mar.  11/76,  Ans.     ^  ^"^  *^^  ^^^^  «^  ^^^^  «*^«' 

item  becoming  due.  After 
which  we  proceed  exactly  as  in  the  Example.  That  is,  we  multiply 
each  item  by  its  time  from  the  focal  date,  and  divide  the  sum  of  these 


AVERAGE     or     PAYMENTS.  379 

products  by  the  sum  of  the  items.  We  find  by  this  process  the  quo- 
tient to  be  37,  which  is  the  number  of  days  after  Feb.  3/76,  when  the 
sum  of  the  items  is  due.  Adding  37  da.  to  Feb.  3/76,  gives  Mar. 
11/76,  Ans. 

Note. — Always  use  the  exact  number  of  days  as  found  in  Table, 
page  237. 

Find  the  average  time  of  payments  of  the  following 
purchases : 

(4.) 
Jan.  6,  $450,  (3  mo.  cr.)        Mar.  24,  $600,  (3  mo.  cr.) 
Feb.  18,  $800,  (3  mo.  cr.)     Apr.  25,  $700,  (3  mo.  cr.) 
Ans.  June  7. 
(5.) 
Apr.  27,  $900,  (6  mo.  cr.)    June  29,  $1000,  (4  mo.  cr.) 
May  16,  $700,  (5  mo.  cr.)     July  14,  $1500,  (3  mo.  cr.) 
Ans.  Oct.  21. 
(6.) 
July  31,  $1200,  (5  mo.  cr.)     Sept.  5,  $8000,  (3  mo.  cr.) 
Aug.  15,  $5000,  (4  mo.  cr.)     Oct.  10,  $6000,  (2  mo.  cr.) 
Ans.  Dec.  11. 

(7.) 
Jan.  2,  mdse.  $1500,  (4  mo.  cr.)     Feb.  2,  mdse.  $2000, 
(4  mo.  cr.)     Jan.  19,  mdse.  $2500,  (4  mo.  cr.)     Mar.  2,  cash 
$3000.  Ans.  Apr.  23. 

(8.) 
B  owes  A,  as  follows :  Mar.  1,  1875,  for  cash,  $400  ;  Apr. 
1,  mdse.  $300,  (3  mo.  cr.) ;  June  1,  mdse.  $500,  (3  mo.  cr.). 
Mnd  the  average  time  for  B's  debts.  Ans.  June  16. 

(9.) 
I  buy  goods,  Jan.  6,  (3  mo.  cr.)  for  $700 ;  Feb.  18,  (4  mo. 
cr.)  for  $800;  Mar.  7,  (3  mo.  cr.)  for  $9000;  Apr.  7,  (1  mo. 
cr.)  for  $1600.     Find  the  average  time  of  payment. 

Ans.  May  31. 


380 


PERCENTAGE. 


Compound  Average, 

Example. — Find  the  equated  time  for  paying  the  bal- 
ance of  the  following  account : 

Dr.      James  Morrow  in  acct.  with  John  Swan.     Or. 


1875. 
Dec.  6. 

1876. 
Jan.  7. 
Feb.  2. 


1876. 

To  Mdse,  4  mo. 

500 

Feb.  1. 
Mar.  3. 

"      "      3  mo. 

600 

Apr.  8.. 

"      **      3  mo. 

400 

By  Mdse,  4  mo. 

600 

"      "      5  mo. 

300 

"  Cash, 

100 

Solution. 

Focal  Date,  Apr.  6,  1876. 

Apr.  6/76,  $500  x  0  =        June  1/76,  $600  x  56  =  33600 

Apr.  7/76,  $600  x  1  =  600    Aug.  3/76,  $300  x  119  =  35700 

May  2/76,  $400  x  26  =  10400    Apr.  8/76,  $100  x   2  =  200 

1500       11000 


1000 


69500 


1500 
1000 


69500 
11000 


500  )  58500 
Apr.  6/76  — 117  da 


117 
Dec.  11/75,  Ana. 


Explanation. 
We  first  select  the  earliest  date  on  either  side  as  the  focal  date,  and 
then  proceed  as  in  Problem  3,  Simple  Average,  to  find  the  product  of 
each  item  by  its  time  from  this  focal  date.  We  next  find  the  sum  of 
these  products  on  the  Dr.  and  Or.  sides,  respectively.  In  this  Ex.  we 
find  the  sum  of  the  products  on  the  Dr.  side  11000,  and  on  the  Cr. 
side  69500  ;  and  the  difference  of  these  sums  58500  We  interpret 
the  result  in  this  way, — the  Dr.  side  is  entitled  to  the  interest  of  $1 
for  11000  days ;  and  the  Gr.  side,  to  the  interest  of  $1  for  69500  days ; 
that  is,  the  Cr.  side  is  entitled  to  the  interest  of  $1  for  58500  (69500  — 
11000)  more  days  than  the  Dr.  side.  The  balance  of  items  on  the  Dr. 
side  is  $500  ($1500  —  $1000).     Therefore,  this  $500  should  have  been 


AVERAGE    OF    PAYME1n[TS. 


381 


paid  enough  earlier  than  the  focal  date,  to  balance  the  interest  of  $1 
for  58500  days  on  the  Cr.  side  ;  or  58500  -j-  500,  or  117  days  before  the 
focal  date,  which  gives  Dec.  11/75,  as  the  equated  time  for  paying  the 
balance  of  the  account. 

We  have  used  the  earliest  as  the  focal  date ;  the  latest  would  be 
just  as  convenient. 

If  an  account  is  settled  at  the  equated  time,  the  balance  of  items  is 
the  true  balance  of  the  account ;  if  hefore  the  equated  time,  the  bal- 
ance of  items  is  discounted  for  the  intervening  time;  if  after  the 
equated  time,  interest  is  allowed  on  the  balance  of  items  from  that 
time  to  settlement. 

Bank  Discount  is  used  without  days  of  grace. 

485.  EuLE. — Multiply  each  debt  hy  the  units  of  time  he- 
tween  its  maturity  and  the  focal  date,  and  divide  the  balance 
of  products  hy  the  balance  of  items.  Then  add  the  number 
of  units  of  time  i7i  the  quotient  to  the  focal  date,  when  both 
balances  are  on  the  same  side  of  the  account,  but  subtract 
from  the  focal  date,  when  balances  are  on  different  sides  of 
the  account. 

Note  —If  the  last  date  is  selected  as  the  focal  date,  we  subtract 

from  the  focal  date  t?ie  number  of  units  of  time  in  the  quotient,  when 
both  balances  are  on  the  same  side  of  the  account ;  and  add  th^m  when 
balances  are  on  different  sides  of  the  account. 

PBOB  LEM  S. 


De. 


(1.) 

James  Johnston". 


Or. 


1875. 

1875. 

Apr.  1. 

To  Cash, 

1200 

Feb.  8. 

By  Mdse,  4  mo. 

2300 

"    21. 

"  Mdse,4mo. 

4000 

Mar.  8. 

«       ((        « 

1500 

May  30. 

«        <(        <( 

3000 

"    28. 

U                   *€                     tt 

1000 

Find  the  face  of  the  note,  and  the  date  from  which  it 
bears  interest,  to  settle  the  above  account. 

Ans.  Face,  $3400;  Date,  Oct.  22/75. 


382 


PEliCEli!  TAGE. 


(3.) 

Equate  the  following  acct.: 

De.    Andeew  Kichey  in  acct.  with  Alex.  Maetin^.    Ce. 


1875. 
Apr.  35. 
May  16. 
June  15. 


1875. 

To  Mdse,  3  mo. 

1700 

May  10. 

U              ((                   it 

2500 

June  3. 

"     •*       <♦ 

5000 

1 

"    27. 

By  Mdse,  3  mo. 
"  Casli, 


2000 
4000 
8000 


Ans.  Martin  owes  $4800,  and  int.  from  May  14/75. 


(3.) 
Balance  the  following  acct.  at  6%,  Sept.  1,  1875 : 
De.        Chas.  Atwell  in  acct.  with  M.  F.  Millee. 


Ce. 


1875. 

1875. 

May  13. 

To  Mdse,  4  mo. 

3560 

40 

June  24 

By  Cash, 

2500 

June  5. 

U                   (C                         t( 

1775 

25  ! 

July  14 

"   Mdse,  4  mo. 

3335 

50 

July  19. 

(t               It                     €( 

2188 

80 

Aug.  1. 

((       ti        (< 

4516 

10 

Ans.  Miller  owes  $2827.15  -$34.87  (dis.  from  Nov.  14) 
=  $2792.28. 

Instead  of  using  Notes  and  Rtjle  on  page  381,  some  accountants 
find  the  value  of  each  item  at  the  required  date.  Thus,  in  Ex.  3,  the 
worth  on  September  1.  1875,  at  6%  (interest  or  discount  depending  on 
the  date,  and  using  Bank  Discount  without  grace) 


Of  .S3560.40  is  $3553.28. 
Of  $1775.25  is  $1765.19. 
Of  $2188.80  is  $2159.98. 

$7478.45. 


Of  $2500.00  is  $2528.75. 

Of  $3335.50  is  $3294.36. 

Of  $4516.10  is  $4447.61. 

$10270.72. 


And  $10270.72  -  $7478.45  =  $2792.27,  amount  due  from  Miller,  Sept.  1. 
1875,  a  result  slightly  different  from  that  found  by  the  other  method. 
Solve  in  like  manner  Ex.  2. 


OUTLINE    OF    POWERS    AND    ROOTS. 


487.  A  POWER.  < 


490.  1st  Power. 

4:91.  2d       "      yor  Square. 

492.  Sd        **      ,  or  Cube. 

493.  Index,  or  Exponent. 

494.  Principle. 

495.  Rule. 


488.  INVOLVINQ  A  NUMBER. 

489.  A  ROOT  OF  A  NUMBER. 


497.  Square  Root. 

ROOTS.            J 

498.  Cube  Root. 

499.  Rational  Roots, 

500.  Swrda. 

501.   THE  SIGN. 

k5 

'502.  Extracting. 

0 

503.  Principle. 

504.  JBuie  /or  Integerg. 

505.  Bute  /or  Fractions. 

0  ^ 
> 

*  506.  Triangle. 

SQUARE 
ROOT, 

M 

507.  iJiflr/t<- 

Angled      ^ 

{  Hypotenuse, 

508.  •<  Perpendicular 

509.  Pnnciple. 

g 

Triangles. 

510.  i?«fe«. 

*1 

511.  J.r(f<M  Q/"  Simila 

- 

. 

Ftane  Figures. 

CUBE 
ROOT. 

51J 

5i: 

(    51^ 

i. 
i. 
1. 

Similar  Solids. 

5i 


CHAPTER  VII. 

SECTION    I. 

— ° g)a(a ' — 


I    IM\f©I,lfJ^iTCIM 


^gG'- 


486.  Involution  is  the  process  of  finding  a  power  of 
a  number. 

487.  A  I^ower  of  a  number  is  the  product  obtained 
by  using  that  number  several  times  as  a  factor.  Thus, 
125  is  a  power  of  5,  being  equal  to  5  x  5  x  5. 

488.  Involving  a  number  is  using  it  to  produce  its 
power. 

489.  A  Moot  of  a  number  is  tbe  number  involved. 
Thus,  5  is  a  root  of  25,  of  125,  of  625,  &c. 

Powers  are  named  from  the  number  of  times  the  root  is  used  as 
a  factor. 

490.  The  First  Power  of  a  number  is  the  number 
itself. 

491.  The  Second  Power,  or  Square,  of  a  number 
is  the  product  obtained  by  using  that  number  twice  as  a 
factor.  Thus,  9  is  the  second  power,  or  square,  of  3 ;  be- 
cause 3x3  =  9. 

493.  The  Third  Power,  or  Cuhe,  of  a  number  is 
the  product  obtained  by  using  that  number  three  times  as  a 
factor.  Thus,  8  is  the  third  power,  or  cube,  of  2 ;  because 
2x2x2  =  8. 

\ 


884 


TKVOLUTION^.  385 

The  second  power  is  called  a  square,  because  the  area  of  the  square 
is  the  product  of  two  equal  factors  (210).  The  third  power  is  called 
a  cube,  because  the  volume  of  a  cube  is  the  product  of  three  equal 
factors.    (216.; 

493.  The  Index,  or  JExponent,  of  a  power  is  that 
number  which  indicates  the  degree,  or  name,  of  the  power. 
It  is  placed  at  the  right  of  the  root,  near  the  top.  Thus, 
5  2  indicates  the  square  of  5 ;  (f )  ^  indicates  the  cube  of  |, 
&c.,  &c. 

494.  The  product  of  two  or  more  powers  of  a  number 
is  a  poiver  indicated  by  the  smn  of  the  indices  of  the  factors. 
Thus,  53  X  5*  =  52+*  =  56,  because  (5  x  5)  x  (5  x  5 
x5x  5)  =  5x5x5x5x5x5. 

To  find  any  power  of  a  number  we  have  the 

495.  KuLE. —  Use  the  number  as  a  factor  as  many  times 
as  there  are  units  in  the  number  denoting  the  power. 

PR  O  B  L  EMS. 


1.  22  =  ?  Ans.  4. 

2.  92  =  ?  A?is.  81. 

3.  102  —9  Ans.  100. 

4.  .992  =  ?  A71S.  .9801. 

5.  1002  ^  9  Ans.  10000. 

6.  9992  =  ?  Ans.  998001. 

7.  23  =  ?  Ans.  8. 

8.  9^  =?  Ans.  729. 

9.  103  =  ?  Ans.  1000. 


10.  993  =  ?  Ans.  970299. 
11. 1003  _  ^  Ans.  1000000. 
12.  9993  =  ?  Ans.  997002999. 
13.8142=?  Ans.  662596. 
14.  8I43  =  ?  Ans.  539353144. 
15.8.362  =  ?  Ans.  69.8896. 
16.  8.363  =  ?  Ans.  584  277056. 
17.(^)2  =  ?    Ans.  A. 

18.(4)3  =  ?    Ans.  ^%. 

Prob.  1-6  illustrate  the  principle,  that  the  square  of  a  nvmber 
Ms  twice  as  many  figures  as  the  number  squared,  or  one  less  than 
twice  as  many. 

Prob.  7-12  illustrate  the  principle,  that  the  cube  of  a  number  Ims 
three  times  as  many  figures  as  the  number  cubed,  or  else  one  or  two 
less  than  three  times  the  number. 
25 


SECTION    II 


II  mw&ii^T^&'m 


496.  Evolution  is  the  process  of  finding  the  root  of  a 
number.    It  is  the  converse  of  involution. 

497.  The  Square  Hoot  of  a  number  is  one  of  two 
equal  factors  whose  product  is  that  number.  Thus,  3  is  the 
square  root  of  9  ;  for  3  is  one  of  the  two  equal  factors  of  9. 

498.  The  Cube  Hoot  of  a  number  is  one  of  the  three 
equal  factors  whose  product  is  that  number.  Thus,  2  is  the 
cube  root  of  8 ;  for  2  is  one  of  the  three  equal  factors  of  8.' 

499.  national  roots  are  those  which  can  be  exactly 
obtained.     Thus,  the  square  root  of  16  is  rational. 

500.  Surd  roots,  or  Surds ,  are  roots  which  cannot  be 
exactly  obtained.     Thus,  the  square  root  of  3  is  a  surd. 

501.  The  Sign  of  JEvohition  is  the  symbol  y^, 
which  is  called  the  radical  sign,  and  is  placed  before 
the  number  whose  root  is  indicated.  Thus,  ^4^  or  ^"4^ 
indicates  the  square  root  of  4 ;  ^27^  indicates  the  cube 
root  of  27.    The  root  may  also  be  indicated  by  a  fraction 

placed  on  the  right  and  at  the  top  of  a  number.     Thus,  4^ 

1  3 

indicates  the  square  root  of  4;  8"^,  the  cube  root  of  8;  4% 

the  square  root  of  the  cube  of  4 ;  that  is,  the  numerator  of 
the  fractional  exponent  indicates  the  power,  and  the  de- 
nominator the  7*00^  of  that  power. 

8^6 


SQUARE     ROOT. 


387 


Square  Root. 

502.  Extracting  the  square  root  of  a  number,  is 
resolving  that  number  into  two  equal  factors. 

In  Prob.  1-6,  Involution,  we  have  seen  that  the  square  of  a  num- 
ber contains  twice  as  many  figures  as  the  number  squared,  or  one  less 
than  twice  as  many.     From  this  it  follows  that 

503.  TJie  square  root  of  a  number  consists  of  one-half 
as  many  figures  as  there  are  figures  in  the  poiver,  or  one- 
half  of  one  more  than  the  number  of  figures  in  the  power. 

We  can,  therefore,  determine  at  once,  the  number  of  figures  in  the 
root,  from  the  number  in  the  power. 

Example  1.  —Find  the  square  root  of  529. 

Before  finding  the  root,  let  us  analyze  the  power,  529.  We  can 
readily  see  that  529  =  400  +  120  +  9,  which  is  equal  also  to  20^  + 
(2  X  3  X  20)  +  33  ;  that  is,  it  is  equal  to 
the  square  of  S  tens  +  2  times  2  tens  x  3 
units  +  the  square  of  3  units. 

Let  us  now  suppose  that  529  means 
529  bricks, each  afoot  square, and  that 
we  wish  to  arrange  them  so  that  they 
will  form  a  square.  In  the  first  pla  'e,  we 
take  the  square  of  the  tens,  that  is  ''/O^, 
or  400,  and  we  have  the  square,  Fig.  1, 
which  is  20  ft.  each  way  and  contains 
400  sq.  ft.  (210). 

Taking  400  from  529  leaves  2  times  2 
tens  X  3  units  -f  the  square  of  the  2  units, 
or  129  sq.  ft.  We  are  not  supposed  as 
yet  to  know  how  many  units  there  are  ; 
but  we  do  know,  that  to  make  our  square 
larger,  we  must  add  a  piece  20  ft.  long 
to  each  of  two  sides  -  equal  to  one  piece 
2  X  20,  or  40  ft.  long.  If  we  are  to  add 
40  ft.  in  length  we  can  find  the  width  by 
dividing  our  material  by  the  length,  40  ft. 
Dividing  129  by  40  we  have  3  for  a  quo- 


388 


EVOLUTION, 


tient  and  a  remainder  9.  This  shows  that  the  two  20  ft.  pieces  are  each 
3  ft.  wide.  Adding  these  two  pieces  (Fig.  2),  we  find  that  we  have 
added  3  x  20=60  to  each  of  the  two  sides  of  our  original  square.  This, 
however,  leaves  a  small  square  of  3  ft.  each  way  to  fill  out.  This 
square  will  take  exactly  the  remaining  9  ft.  ;  and  we  have  a  square 
23  ft.  each  way,  composed  of  20^  +  (2  x  3"  x  20)  +  3^.  Had  there 
been  a  remainder,  we  could,  in  like  manner,  have  added  to  each  of  two 
sides,  a  piece  23  ft.  long,  whose  width  woula  have  been  detenxmxdd 
by  dividing  2  x  23  into  the  remainder. 


SOLUTION. 


Power,  529 

4 


23  Root. 


43 


129 
129 


Explanation.  —  We  separate  the 
number  beginning  at  the  right  into 
the  periods  29  and  5.  As  5  contains 
the  square  of  two,  the  tens  must  be 
two.  We  square  two  tens,  and  subtract 
the  square  from  5.  This  leaves  1,  to 
which  we  annex  29,  giving  us  a  re- 
mainder of  129.  We  then  double  the 
two  tens,  or  20  unite,  as  we  have  to  add  to  two  sides ;  by  this  we  divide 
129,  and  obtain  for  a  quotient,  3,  which  we  add  to  the  20  and  the  40, 
(or  which  is  the  same  thing  annex  it  to  the  2  and  4),  and  then  multi- 
plying 40  +  3  =  43,  by  3,  the  second  figure  of  the  root,  we  obtain  129, 
which  subtracted  from  129  leaves  nothing,  and  our  work  is  complete. 

Example  2. — What  is  the  square  root  of  801025  ? 


solution. 

801025  I  895 
64 


169 


Explanation, — We  first  find  two 
figures  of  the  root  as  in  Ex.  1.  Then  to 
our  remainder  we  annex  the  next  period 
of  the  power  for  a  new  dividend.  For  a 
divisor,  we  double  the  root  already  found, 
89,  (for  we  still  have  two  sides  to  in- 
crease), giving  us  178,  and  find  how  often 
it,  178,  is  contained  in  the  dividend,  ex- 
clusive of  the  right-hand  figure.  The 
result  we  annex  to  both  divisor  and  root 
already  found,  and  then  multiply  the  divisor  thus  increased  by  the 
last  figure  of  the  root,  and  subtract  the  product  from  the  dividend. 
Result,  895. 


1610 
1521 


1785 


8925 
8925 


5852.25 
49 

76.5 

46   952 
876 

1525 

76  25 
76  25 

SQUARE     ROOT.  389 

Example  3. — Find  the  square  root  of  5852.25. 

SOLUTION.  „  ^,.  ,^,, 

^^^^  „^^  Explanation.— Since  lOths  squared 

produce    hundredths,   and    hundredths 

squared  produce    ten-thousandths,   the 

square  rootof  hundredths  must  be  tenths, 

and  the  square  root  of  ten-thousandths 

must  be  hundredths.    Therefore,  if  we 

have  a   decimal  whose  square  root   is 

required,  we  must  point  off  from  the 

decimal  point  to  the  right,  and  must 

have,  in   pairs,  tenths  and  hundredths, 

thousandths  and  ten-thousandths,  &c. ;  and  if  any  place  or  period  i 

wanting  it  must  be  supplied  with  a  cipher  or  ciphers  :  thus,  if  we  wish 

the  square  root  of  35.7  we  should  write  25.70 ;  of  325. 675,  ^325. 6750,  &c. 

504.  Rule. — Begin  at  the  decimal  point  and  separate  the 
power  into  periods  of  two  figures  each. 

Find  the  greatest  square  in  the  left-hand  period,  and  write 
its  root  as  a  quotient  in  division. 

Suhtract  the  square  of  this  figure  from  the  left-hand  period, 
and  to  the  remainder  annex  the  next  period. 

Use  this  result  as  a  dividend,  and  for  a  divisor  double  the 
figure  in  the  root.  See  how  often  this  divisor  is  contained  in 
the  dividend,  exclusive  of  the  right-hand  figiire,  and  twite  the 
quotient  to  the  right  of  the  divisor  and  also  to-the  right  of  the 
first  figure  of  the  root. 

Multiply  the  resulting  divisor  by  the  last  figure  of  the 
root,  a7id  subtract  the  product  from  the  dividend. 

To  the  remainder  annex  the  next  period  for  a  new  divi- 
dend ;  and  for  a  divisor  double  the  root  already  found,  and 
proceed  as  before,  till  all  the  periods  are  used,  or  until  the 
root  is  obtained  with  sufficient  accuracy. 

When  a  divisor  is  not  contained  in  its  dividend,  exclusive  of  the 
right-hand  figure,  annex  a  cipher  both  to  the  divisor  and  root,  bring 
down  the  next  period,  and  proceed  as  before. 


390 


EVOLUTION. 


ritOBLEMS. 

Find  the  square  root 

1. 

Of  625. 

Ans.     25. 

6.  Of  260. 

Ans. 

16.1245. 

2. 

Of  6.25. 

Ans.    2.5. 

7.  Of  347. 

Ans. 

18.6279. 

3. 

Of  63001. 

Ans.  251. 

8.  Of  1020. 

Ans. 

31.9374 

4.    Of  64009.      Ans.  253. 


9.  Of  10.45.      Ans.     3.2326. 


5.   Of  6.4009.     Ans.  2.53.     10.  Of  105400.  Ans.  324.6536. 

To  find  the  Square  Hoot  of  a  Fraction. 

505.  KuLE. — Divide  the  square  root  of  the  numerator 
hy  the  square  root  of  the  denominator. 

The  reason  for  this  is  plain  when  we  consider  that  the  square  of  a 
fraction  is  the  sqtiare  of  its  numerator  divided  by  the  square  of  the 
denominator. 

Find  the  square  root 

14.  Of    m-     ^^s.    «. 

15.  Of  ^^Vr-  ^^^-    VW- 

16.  Of     A.      Ans.■L^^^^. 


11.   Of  A- 

Ans. 

h 

12.   Of-^. 

Ans. 

tV 

13.   Of  J|. 

Ans. 

i- 

Applications  of  Square  Moot. 

506.  A  plane  figure  bounded  by  3 
sides  is  called  a  Triangle. 

507.  When  one  angle  of  a  triangle 
is  a  right  angle  (206),  the  figure  is 
called  a  Might-angled  Triangle. 
Thus,  in  the  figure  ABC,  B  is  a  right 
angle,  and  ABC  is  a  right-angled  tri- 
angle. 

508.  The  side  AC,  opposite  the  right 
angle,  is  called  the  Hypotenuse.  The  other  two  sides, 
BC  and  AB,  are  called  the  Base  and  Perpendicular, 
respectively. 


APPLICATIONS     OF     SQUARE     ROOT, 


391 


All  right-angled  triangles  possess  this  property,  namely: 
509.    Tlie  square  described  on  the  hypotenuse  is  equivor 
lent  to  the  sum  of  the  squares  on  the  other  two  sides. 

This  principle  is  illus- 
trated by  Fig.  1,  in 
which  it  is  readily  seen 
that  the  9  +  the  16 
small  squares  of  the 
squares  described  on  the 
two  sides  are  equivalent 
to  the  25  small  squares 
of  the  square  on  the 
hypotenuse. 

We  see,  too,  that  if 
we  know  the  number  of 
squares  on  two  of  the 
sides  we  can  readily  find 
the  number  on  the  other 
side.  For  9  +  16  =  25  ; 
or,  25-  16  =  9;  or,  25 
—  9  =  16.  In  like  man- 
ner, if  the  sides  were  respectively  5,  12,  and  13,  then  5^+12^=  13^,  or 
132-123=52,  or  132-52=122.     Hence,  the 

510.  Utiles. 

To  find  the  Hypotenuse, 
Extract  the  square  root  of  the  sum  of  the  squares  of  the 
other  tivo  sides. 

To  find  the  Base  or  Perpendicular, 
Extract  the  square  root  of  the  difference  of  the  squares 
of  the  hypotenuse  and  the  given  side. 

PR  OB  L  EMS, 

Find  the  hypotenuse  when  the 

1.  Base  is  30,  perpendicular  40  ft.  Ans.  50  ft. 

2.  Base  is  24,  perpendicular  10  ft.  Ans.  26  ft. 

3.  Perpendicular  is  60,  base  80  ft.  A71S.  100  ft. 

4.  Perpendicular  is  40,  base  96  ft.  Ans.  104  fi 


392  EVOLUTION. 

Find  the  other  side  when 

5.  One  side  is  20,  hjrpotenuse  52.  Ans.  48. 

6.  One  side  is  14,  hypotenuse  17^.  Ans.  10^. 

7.  How  much  farther  half  way  around  a  square  ten-acre 
field  than  along  its  diagonal  ?  Ans.  23.432  rods. 

8.  How  long  must  a  rope  be  that  will  reach  from  the  top 
of  a  house  40  ft.  high  to  the  bottom  of  a  house  on  the  other 
side  of  a  60  ft.  street  ?  Ans.  72.11. 

9.  The  width  of  a  house  is  24  ft.  and  the  height  of  the 
comb  is  6  ft.  above  the  walls.  What  is  the  length  of  the 
rafters,  allowing  2  ft.  for  cornice  ?  Ans.  15.416  ft. 

10.  What  is  the  longest  inflexible  ruler  that  can  be  placed 
in  a  box  8  X  10  X  15  inches?  Ans.  19.723  in. 

The  following  truth  is  fully  demonstrated  in  Geometry ;  we  can 
only  state  it  in  this  place. 

511,  The  areas  of  all  similar  plane  figures  are  to  each 
other  as  the  squares  of  the  like  dimensions  of  the  figures. 

A  square  whose  side  is  twice  as  long  as  that  of  another  has  four 
times  the  area  ;  a  circle  whose  diameter  is  three  times  that  of  another 
has  nine  times  the  area. 

11.  A's  lot  is  20  rd.  square;  B's,  60  rd.  square.  How 
many  lots  the  size  of  A's  equals  B's  ? 

Solution.— 602  =  3600  ;  20^  =  400  ;  3600  -=-  400  =  9,  Ans. 

12.  How  many  times  as  large  is  a  circle  8  in.  in  diameter, 

as  one  4  in.  in  diameter  ?  Ans.  -—=-—  =  4. 

4^        10 

13.  How  many  times  as  much  surface  in  a  circle  5  in.  in 
diameter,  as  in  one  1  in.  in  diameter  ?  Ans.  25. 

14.  How  many  times  as  much  water  will  pass  through  a 
circular  aperture  lOJ  in.  in  diameter,  as  one  3^  in.  in  diam- 
eter? Ans.  (10i)2  -f-  (3i)2  =  9^3^. 


APPLICATIONS     OF     SQUARE     ROOT.  393 

15.  How  do  the  faces  of  two  clocks  compare,  when  one  is 
3  in.  in  radius  and  the  other  4  in.  ? 

Ans.  The  latter  =  If  times  the  former. 

16.  I  have  two  plots  of  land  exactly  alike  in  shape,  and  a 
diagonal  of  one  is  25  rd.,  while  the  corresponding  diagonal 
of  the  other  is  75  rd.     What  are  their  relative  sizes  ? 

Ans.  The  latter  =  9  times  the  former. 

17.  Of  two  lots  exactly  alike  as  to  shape,  one  requires  44 
rd.  of  fence,  and  the  other  88  rd.  How  many  times  the 
former  equals  the  latter  ?  Ans.  4  times. 

18.  I  have  a  rectangular  lot  40  rods  long  and  4  rods  wide. 
What  are  the  dimensions  of  a  lot  of  the  same  shape  con- 
taining 2  acres  ?  Ans.  5.6568  x  56.668  rods. 

19.  One  of  the  same  shape  containing  3  acres  ? 

.     Solution.— 1  acre  :  3  acres  : :  4^  :  (6.928)^ 

: :  402  :  (69.28)2. 

CUBJS   HOOT. 

In  Proh.  7-12  of  Involution,  we  have  seen  that  the  cube 
of  a  number  has  three  times  as  many  figures  as  the  number, 
or  else  one  or  two  less  than  three  times  as  many.  Hence  it 
follows  that 

513.  Tlie  Cuhe  root  of  a  number  has  1  figure  for  every 
group  of  three  figures  in  the  power,  and  1  figure  for  any 
excess  over  any  number  of  groups  of  three  figures  each  in 
the  power. 

We  can,  therefore,  determine  by  inspection  the  number  of 
figures  in  the  cube  root  of  any  power. 

Example  1. — Find  the  cube  root  of  12167. 

Let  us  analyze  12167.  We  find  that  it  consists  of  8000  +  (3  x  400 
X  3)  +  (3  X  20  X  9)  +  27  =  20^  +  (3  x  20=^  x  3)  +  (3  x  20  x  3^)  +  33. 
Or,  to  put  the  expression  in  another  form,  it  equals  the  cube  of  2  tens 


394 


EVOLUTlOi^^. 


+  3  times  the  square  of  2  tens  x 
of  3  units  +  the  cube  of  3  units. 

Fig.  1 


Fig.  3. 


units  +  3  times  2  tens  x  the  square 


Let  12167  rep- 
resent cu.  ft.,  or 
tjie  volume  of  a 
cube  whose  di- 
mensions we  wish 
to  find.  The 
inner  block  in 
Fig.  1  represents 
a  cube  each  of 
whose  dimen- 
sions is  20  ft.  and 
whose  volume  is 
20  X  20  X  20  = 
(20=^)  8000  cu.  ft. 
Deducting  8000 
cu.  ft.  from  12167 
cu.  ft.  we  have 
4167  cu.  ft.;  which 
is  the  volume  of 
the  material  to 
be  applied  to  the 
cube  in  Fig.  1. 
In  order  to  en- 
large a  cube  and 
yet  preserve  its 
form,  the  addi- 
tions must  be 
made,  first,  to 
three  adjacent 
sides,  as  the  top, 
front,  and  right ; 
second,  to  three 
edges ;  and  third, 
to  one  comer. 
The  three  rectan- 
gular slabs  about 
the  cube  in  Fig. 
1  represent  the 
parts    to   be  ap- 


CUBE     ROOT, 


395 


Fig.  3.  plied    to  the   adjacent 

sides  of  tlie  cube.  Each 
slab  is  20  ft,  in  length 
and  width  and  contains 
a  mrfoM  of  30  x  20  = 
(202)  400  sq.  ft.  The 
three  contain  3  x  20 
X  20  =  3  X  203  = 
1200  sq.  ft.  The  thick- 
ness of  these  side  slabs 
we  determine  by  divid- 
ing the  number  repre- 
senting the  volume  of 
the  material  to  be  ap- 
plied by  the  number 
representing  the  surface 
of  the  side  slabs.  That  is,  we  divide  4167  by  1200,  thus  obtaining  a 
quotient  3  which  represents  the  thickness  of  each  slab.  The  volume 
of  each  of  these  slabs  is  20  x  20  x  3  =  20^  x  3  =  1200  cu.  ft.  The 
volume  of  the  three  is  3  x  20  x  20  x  3  =  3  x  20^  x  3  =  3600  cu. 
ft.  Fig.  2  shows  the  form  of  the  solid  after  the  three  rectangular 
slabs  have  been  applied.    These  additions,  however,  leave  our  cube 

incomplete,  as    there 


Fig.  4. 


are  three  edges  to  be 
filled  by  the  blocks 
marked  C,  C,  C,  in 
Fig.  3.  Each  of  these 
blocks  is  20  ft.  long, 
and  in  thickness  and 
width  the  same  as  the 
thickness  of  the  rect- 
angular slabs,  that  is 
3  ft.  The  volume  of 
each  is.  therefore,  3  x 
3  x  20  =  32  X  20  = 
180  cu.  ft.,  and  the 
volume  of  the  three 
blocks  is  3  X  3  X  3 
X  20  =  3  X  32  X  20 
=  540  cu.  ft.    Fig.  4 


396 


EVOLUTION. 


cube  as 
1. 
2. 

3. 

4 


Fig.  5.  represents  the  form  of 

tlie  solid  after  tlie  three 
edge  blocks  have  been 
placed  in  their  proper 
places.  We  observe 
that  our  cube  is  still 
incomplete,  as  there  U 
one  corner  to  be  filled 
by  the  block  marked  D 
in  Fig.  3.  Each  of  the 
dimensions  of  this  block 
is  evidently  3  ft.,  and 
hence  its  volume  is  3  x 
3  X  3  =  38  =  27  cu.  ft. 
Placing  this  block  in 
its  proper  position  in 
Fig.  4  we  have  oar 
complete  and  enlarged 

represented  by  Fig.  5.    This  cube  is  composed  of 

A  cube  whose  edge  is  20  ft.  =  20^  =     .    .  8000  cu.  ft. 

Three  rectangular  slabs  each  20    x   20  x  3 ; 

or  3  X  202  X  3  = 3600  cu.  ft. 

Three  rectangular  blocks  each  20  x  3  x  3 ; 

or  3  X  20  X  32  = 540  cu.  ft. 

A  cube  whose  edge  is  3  ;  or  3  x  3  x  3  =  3^  =      27  cu.  ft. 

12167  cu.  ft. 


12167  (  23 
8 


SOLUTION.  Explanation. — We  first 

separate  the  power  into  two 
periods,  12  and  167. 

Since  the  cube  of  units 
produces  units,  tens,  or  hun- 
dreds, and  the  cube  of  tens 
can  produce  no  lower  order 
than  thoTisands,  the  cube  root 
of  12167  must  be  units  and 
tens,  or  2  figures ;  and  the 
first  figure,  or  tens,  must  be  the  cube  root  of  some  cube  in  12.  8  is 
the  largest  exact  cube  in  12,  and  2  is  its  cube  root ;  and  we  conclude 
that  2,  or  2  tens,  is  the  first  figure  of  the  root.  Subtracting  the  cube 
of  2  tens  from  12  and  bringing  down  the  next  period,  or  what  is  the 


2«  X  300  =  1200 

2  X  30  X  3  =  180 

32  =   9 

4167 

1389 

4167 

CUBE     ROOT. 


397 


game  thing,  subtracting  the  cube  of  20,  8000,  from  12167,  we  have  left 
4167.  This  is  to  be  placed  on  3  faces,  but  these  faces  are  each  20  ft. 
square.  They  have,  therefore,  20  x  20  x  3,  or  1200  ft.  of  surface.  Di- 
viding 4167  by  1200,  the  number  of  ft.  in  this  surface,  we  obtain  its 
thickness,  which  is  at  least  3  ft.  But  we  must  make  provision  also 
for  the  three  edge-pieces  which  are  each  20  x  3^,  and  the  corner  cube 
which  is  3^.  These  seven  parts  are  equal  in  volume  to  (20^  x3)x3  + 
(20  X  3  X  3) X 5-f-(3)2  X 5=  (208  X  ;} +  20  X  3  X  3-f-32)  X 5=  (1200  +  180  +  9 
=  1389) x3  =  4167 ;  this  subtracted  from  4167  leaves  no  remainder, 
and  the  work  is  complete. 

Example  2.— Extract  the  cube  root  of  12812904. 


SOLUTION. 


12812904  ( 234. 
8 


Explanation. — We 
obtain  the  first  two 
figures  of  the  root  as 
in  Ex.  1.  Then  hav- 
ing annexed  to  the 
remainder  the  last 
period  for  a  new  divi- 
dend, we  take  300 
times  the  square  of 
the  root  already 
found,  that  is,  the 
faces  of  the  cube  are 
now  230  ft.  square, 
and  there  are  3  of 
them  (2302  ^  3=232  ^ 
300),  and  by  division 
we  find  their  thickness  4.  We  then  add  to  23^  x  300,  23  x  30  x  4  (=230 
X  3  X  4),  for  the  length  and  breadth  of  the  three-edge  pieces ;  and  add 
also  4^  for  length  and  breadth  of  corner  cube ;  and  then  multiply  the 
sum  by  4,  the  thickness  ;  and  since  the  product  is  645904,  there  is  no 
remainder,  and  the  work  is  complete. 

Since  the  cube  of  tenths  produces  thousandtTis,  and  the  cube  of  hun- 
dredtJis,  millionths,  when  we  extract  the  cube  root  of  a  decimal,  that 
decimal  must  have  the  form  of  thousandths,  mUlionths,  &c.  Thus,  the 
cube  root  of  3.5  equals  the  cube  root  of  ?.500,  orUsOOOOO,  &c. 

513.  EuLE. — Separate  the  poiver  into  periods  of  three 
figures  each,  beginning  at  the  decimal  point. 


28  X  300  =  1200 

4812 

2  X  30  X  3  =  180 

32  =        9 

1389 

4167 

232  X  300  =r  158700 

645904 

23  X  30  X  4  =   2760 

42  =    16 

161476 

645904 

398  •     EVOLUTION. 

For  the  first  figure  of  the  root,  write  the  cube  root  of  the 
greatest  exact  cube  in  the  first,  or  left-hand  period. 

Subtract  the  cube  of  this  first  figure  of  the  root  from  the 
first  period,  and  to  the  remainder  annex  the  next  period. 
Regard  the  result  as  a  dividend. 

For  a  divisor  take  300  times  the  square  of  the  first  figure 
of  the  root.  Find  how  often  the  product  is  contained  in  the 
dividend,  and  write  the  quotient  as  the  second  figure  of  the 
root. 

Complete  the  divisor  by  adding  to  it 

1st.  The  product  of  the  first  figure  of  the  root  by  SO  times 
the  last  figure  of  the  root  ; 

2d.  The  square  of  the  last  figure  of  the  root. 

Multiply  the  divisor  thus  increased  by  the  last  figure  of 
the  root ;  subtract  the  product  from  the  dividend,  and  to  the 
remainder  annex  the  next  period  for  a  neiu  dividend. 

Form,  in  the  same  manner,  successive  divisors,  and  find 
corresponding  figures  of  the  root ;  being  careful  to  complete 
each  divisor  by  the  addition  of  30  times  the  product  of  the 
last  figure  by  the  preceding  figures  of  the  root,  and  also  the 
square  of  the  last  root  figure. 

When  any  product  is  greater  than  the  dividend  erase  the  figure 
that  produced  it,  and  with  a  figure  of  less  value  recalculate  the  addi- 
tions to  the  trial  divisor  until  the  product  is  smaU  enough  for  sub- 
traction. 

When  any  trial  divisor  is  not  contained  in  the  dividend,  annex  a 
cipher  to  the  root,  two  ciphers  to  the  trial  divisor,  bring  down  the 
next  period  to  the  right  of  the  dividend,  and  proceed  as  before. 

The  cube  of  a  common  fraction  is  the  cube  of  the  numerator  writ- 
ten over  the  cube  of  the  denominator ;  therefore,  the  cube  root  of  a 
common  fraction  is  the  cube  root  of  the  numerator  over  the  cube  root  of 
the  denominator. 


CUBE     BOOT. 


399 


rjt  O  B  L  E  MS  . 

Find  the  cube  root  of 


1.  2197. 

2.  9261. 

3.  59319. 

4.  238328. 

9.  .277167808. 


Ans.  13. 
Ans.  21. 
Ans.  39. 
Ans.  62. 

Ans. 


5.  1154320649.      Ans.  1049. 

6.  924010.424.       Ans.  97.4. 

7.  633.839779.       Ans.  8.59. 

8.  2.  J?i5.  1.259921. 


.652. 


10.  .000410172407.    Aiis.  .0743. 

11.  .002.  Ans.  .12599. 


12.  Hff. 
1^-  ^^^#ftr. 


Ans.  H. 


514.   It  is  demonstrated  in  Geometry  that 


All  similar  solids  are  to  each  other  as  the  cubes  of  their 
like  dimensions. 

We  make  a  few  applications  of  this  principle,  referring 
the  pupil  to  the  Section  on  Mensuration  for  definitions  of 
pyramids,  cones,  &c. 

PJJ  OBLEM8, 

1.  How  many  times  will  a  cube  whose  edge  is  3  in.  con- 
tain one  whose  edge  is  1  in.  ?  Ans.  27  times. 

2.  If  an  apple  1  in.  in  diameter  is  worth  If,  how  much  is 
one  2  in.  in  diameter  worth  ?  Ans.  8i*. 

3-  A  teamster  who  had  a  cart-bed  4  x  6  x  1  ft.,  made 
another  in  which  each  dimension  was  increased  J.  How 
does  the  former  compare  with  the  latter? 

4.  Of  two  similar  pyramids,  one  has  a  height  of  2  ft.  and 
the  other  of  5  ft.     How  do  they  compare  in  volume  ? 

Ans.  23  :  53. 

5.  A  Winchester  bushel  is  represented  as  a  cylindrical 
vessel  18^  in.  in  diameter,  and  8  in.  deep.  What  would  be 
the  dimensions  of  a  similar  vessel  that  would  hold  2  bushels? 

Ans.  Dia.,  23.308 ;  depth,  10.08  in. 


OUTLINE    OF    PROGRESSIONS. 


a 

O 


616.  Terms. 

617.  Ascending  Series, 

618.  Descending  Series, 

619.  Extremes, 

620.  Means. 


622 

.  To  find  Last  Term. 

623.  To  find  Sum. 

621.  Arithmetical.  ^ 

624.  ToyiTit?  Common  Bif 

3Q 

626.  2'(?  ^wii  NuTnber  of 

1 

Terms. 
^  627.  r^^^wtZ  ^;iy  Term. 

628.  To  find  Sum. 

626.  Geometrical.     ' 

ANNUI- 
TIES. 

'  629.  Definition. 

630.  Amount. 

631.  TofindAmt. 

4011 


CHAPTER    VIM. 


y\ 


F-mOJQKESSlSMS  1: 


^^^ 


515.  A  Series^  or  Progression^  is  a  succession  of 
numbers,  each  of  which  has  the  same  relation  to  its  imme- 
diately preceding  number.  Such  u  umbers  are  said  to  be  in 
progression. 

516.  The  Terms  of  a  Series  are  the  numbers  which 
compose  it. 

517.  An  Ascending  Series  is  one  composed  of  terms 
each  of  which  is  larger  than  the  preceding  term. 

518.  A  Descending  Series  is  one  composed  of  terms 
each  of  which  is  smaller  than  the  preceding  term. 

519.  The  Extremes  of  a  series  are  the  first  term  and 
the  last  term. 

520.  The  Means  of  a  series  are  all  the  terms  except 
the  first  and  last. 

In  reference  to  the  ratio  of  their  terms,  series  are  either  Arith' 

metical  or  OeometricaZ. 

ABITHMETTCAL    SEBIES. 

521.  An  Arithmetical  Series  is  one  whose  terms 
increase,  or  decrease,  by  a  number  called  the  Common 
Difference.  Thus,  1,  4,  7,  10,  &c.,  is  an  asceriding,  and 
25,  22,  19,  16,  13,  &c.,  is  a  descending  series,  in  each  of 
which  the  common  difference  is  3. 


402  PROGRESSIONS. 

To  find  the  Last  Term^  when  tJie  Common 
Difference  and  First  Term  are  given. 

Example. — Find  the  10th  Term  when  the  1st  term  is  2, 
and  the  common  difference  is  3. 

SOLUTION.  Explanation.— It  is  evident 

2     I     /JO  1  ^    X   3  =  29  *^^*  ^^^^  second  term  =  the  first 

term  +  once  the  common  diflfer- 
ence,  and  the  tTmrd  term  =  the  first  term  +  twice  the  common  differ- 
ence, and  so  on.  We  conclude,  therefore,  that  the  tenth  term  is  equal 
to  the  first  term  +  nine  (10  —  1)  times  the  common  difference ;  and  in 
general  that 

522.  Rule.  —  Any  term  equals  the  first  term  plus  as 
many  times  the  common  difference  as  there  are  units  in  the 
number  of  the  term  less  one. 

If  the  series  is  a  descending  one,  we  use  the  word  minus  for  plus  in 
the  Rule. 

mOBZEMS. 

1.  Find  the  15th  term  of  3,  5,  7,  &c,  Ans.  31. 

2.  Find  the  30th  term  of  1,  3,  5,  7,  &c.  Ans.  59. 

3.  Find  the  16th  term  of  1J-,  3,  4|,  &c.  Ans.  24. 

4.  Find  the  20th  term  of  1,  IJ,  IJ,  1},  &c.        Ans.  5|. 

5.  If  at  6%  per  annum  the  amount  of  $400  is  1424,  $448, 
&c.,  what  is  the  amt.  at  the  end  of  the  11th  year?  (No.  of 
terms  12.)  Ans.  $664. 

To  find  the  Sum  of  a  Series. 

Example. — Find  the  sum  of  5  terms  of  the  series  7, 10, 

13,  &c. 

SOLUTION.  Explanation.  — 

7  _|.  10  +  13  -h  16  +  19  =  Sum.  Writing  the  terms  of 

ci  *li®    series    in    their 

19  +  16  +  134-10+    7  =  Sum.  ^^,^^^^  ^,^,,^  ,^^^ 

26  +  26  +  26  +  26  +  26  =  2  Sums,  or      taking  them   in  an 

5  X  26 ;  i^  =  65,  Sum.  ^/^^"^^  ^'^^''   """f 

*  adding  them  togetn- 


AEITHMETICAL     SERIES.  403 

er,  we  have  twice  the  sum  of  the  series  =  5  x  26  =  5  x  sum  of  the 
extremes  (or  of  any  two  terms  equidistant  from  the  extremes),  which 
divided  by  2  gives  once  the  sum,  65.     We  have,  therefore,  this 

523,  Rule. — 77ie  sum  of  a  series  is  equal  to  one-half  the 
sum  of  the  extremes  multiplied  hy  the  number  of  terms, 

PMOB1.EMS. 

1.  Find  the  sum  of  10  terms  of  the  series  5,  8,  11,  &c. 
(Find  Last  Term  by  522.)  Ans.  185. 

%.  Find  the  sum  of  16  terms  of  the  series  4,  8,  12,  &c. 

Ans.  544. 

3.  Find  the  sum  of  12  terms  of  the  series  $6,  $12,  118, 
&c.  Ans.  $468. 

4.  Find  the  sum  of  100  terms  of  the  series  -J,  f ,  1,  &c. 

Ans.  1287}. 

To  find  the  Co f union  Difference,  when  the 
dumber  of  Terms  and  Extremes  are  known. 

Example. — What  is  the  common  difference  when  the  first 
term  is  3,  and  the  22d  term  is  45  ? 

Solution. — Since  the  last  term  is  equal  to  the  first  term  plus  the 
common  difference  multiplied  by  the  number  of  terms  less  one,  if  we 
subtract  3  from  45,  we  must  have  left  the  common  difference  multi- 
plied by  the  number  of  terras  less  one. 

The  number  of  terms  less  one  is  21 ;  and  42  (=  45  —  3)  divided  by 
21  gives  2,  the  common  difference. 

524.  Rule. — Divide  the  difference  of  the  extremes  hy  the 
number  of  terms  less  one. 

¥RO  BT.  E  M  S. 

What  is  the  common  difference  when 

1.  The  first  term  is  1,  and  the  21st,  41  ?  Ans.  2. 

2.  The  first  term  is  93,  and  the  32d,  0  ?  Ans.  3. 

3.  The  first  term  is  38,  and  the  21st,  28  ?  Ans.  f 


404  progressio:n^s. 

4.  The  amount  of  $400  at  simple  int.  for  12  yr.  is 
"What  is  the  rate  ^  ?  Ans.  6%. 

Although  the  time  is  12  yr.  the  number  of  terms  is  13. 

To  find  the  Number  of  Terms  when  the  Ex- 
tremes and  Common  Difference  are  known. 

The  difference  of  the  extremes  =  common  difference  x  No.  of  terms 
less  1.     (Ex.  and  Solution,  p.  403.)     Therefore, 

52i5.  Rule. — Add  1  to  the  quotient  of  the  difference  of 
the  extremes  divided  hy  the  cotmnon  difference, 

PROBLEMS. 

What  term  of  the 

1.  Series  3,  6,  9,  &c.  is  27  ?  Ans.  9th. 

2.  Series  6,  8,  10,  12,  &c.  is  48  ?  Ans.  22d. 

3.  Series  27,  24,  21,  &c.  is  0?  Ans.  10th. 

4.  Series  |,  },  J,  1,  &c.  is  28?  Ans.  220th. 

GE03TETBICAL    SERIES. 

526.  A  Geometrical  Series  is  a  series  increasing, 

or  decreasing,  by  a  constant  multipHer,  called  the  ratio. 

The  series  is  ascending ,  when  the  ratio  is  greater  than  unity ; 
and  descending f  when  the  ratio  is  less  than  unity.  Thus,  |,  1,  3,  4, 
8,  &c.  is  ascending,  the  ratio  being  2 ;  and  8,  4,  2,  1,,  |,  &c.  is  descending, 
the  ratio  being  ^. 

To  find  any  required  Term^  when  the  First 
Term  and  the  Hatio  are  known. 

Example.— Find  the  7th  term  of '2,  6,  18,  &c. 

SOLUTION.  Explanation.— The  second  term  is  the  first 

3^  =  729  multiplied  by  the  first  power  of  the  ratio;   the 

2    X  729  =^  1458    ^^^*^^  ^^^^  ^^  *^^^  ^^®^  multiplied  by  the  second 

poicer  of  the  ratio,  &c.     Hence,  the  7t?i  term  is 

the  first  muldplied  by  the  6th  power  of  the  ratio. 


GEOMETRICAL     SERIES.  405 

5211,  EuLE. — Multiply  the  first  term  hy  that  poive'^  of  the 

"^atio,  whose  exponent  is  one  less  than  the  number  of  the 
required  term. 

PR  O  BJj  EMS. 

1.  Find  the  6th  term  of  1,  2,  4,  &c.  Ans.  32. 

2.  Find  the  8th  term  of  3,  6,  12,  &c.  Ans.  384, 

3.  Find  the  9th  term  of  1,  3,  9,  &c.  Ans.  6561. 

4.  Find  the  6th  term  of  4,  2,  1,  &c.  Ans.  \. 

5.  Find  the  5th  term  of  1.06,  1.06^,  1.063,  &c. 

Ans.  1.3382256 +  . 

6.  A  man  bought  a  house  containing  24  windows,  at  3 
mills  for  the  first,  6,  for  the  second,  &c.  What  did  the  last 
window  cost?  ^;^s.  $25165.824. 

To  find  the  Sum  of  a  Geoinetrical  Seines, 

Example.— Find  the  sum  of  the  series  6,  18,  54, 162,  486. 

BOLUTION. 

6  4-  18  +  54  4-  162  +  486  =  Sum  of  the  series. 

18  4-  54  +  162  +  486  +  1458  =  3  times  the  sum. 

1458  —  6  =  1452  =  (3  —  1)  times  the  sum. 
1458  —  6       1452 


3  -  1     "■     2 


726,  Sum  of  the  series. 


Explanation. — We  multiply  each  term  of  the  given  series  by  3, 

the  given  ratio,  and  the  result  is  3  times  the  series.     We  then  subtract 

from  3  times  the  series,  1  time  the  series.     The  difference  must  equal 

2  times  the  series,  and  this  difference  divided  by  2  (the  ratio  less  1) 

(3  X  3^  —  6 
equals  the  sum.     Or,  — - — - —  =  726. 
o  —  1 

52iS,   Rule. — Multiply  the  last  term  by  the  ratio ;  then 
divide  the  difference  between  this  product  and  the  first 
term  by  the  difference  between  the  ratio  and  unity. 
If  a  descending  series  is  infinite,  the  last  term  is  0. 


406  PKOGRESSIONS. 

PMO  BLEMS, 

Find  the  sum 

1.  Of  8  terms  of  5,  15,  45,  &c.  Ans.  16400. 

2.  Of  10  terms  of  1728,  864,  &c.  Ans.  3452f . 

3.  Of  1,  i,  i,  \,  &c.,  to  infinity.  A7is.  2. 

4.  Of  yV  Too»  tA-o>  &c.,  to  infinity.  Ans,  f 

5.  A  man  bought  a  house  containing  24  windows,  at  3 
mills  for  the  first  window,  6  for  the  second,  12  for  the 
third,  &c.    What  did  he  pay  for  the  house  ? 

^/^s.  $50331.645. 

ANNUITIES. 

529.  An  Annuity  is  a  sum  of  money  payable  annually, 
or  at  regular  periods  of  time. 

530.  The  Afnoiint  of  an  annuity  is  the  sum  of  the 

payments  and  their  interest  for  the  specified  time. 

To  find  the  amount  of  an  Annuity  at  Com^ 
pound  Interest. 

Example. — To  what  will  an  annuity  of  $10  amount  in 
3yr.,  at6^? 

SOLUTION.  Explanation. — We  first 

$10  X  1.062  =  $11,236  find  the  amount  of  $10  for 

$10  X  1.06     =  $10.60  ^  ^^-  '  *^^^  ^^  ^^^  ^^^  1  y^- ' 

$10  =  $10.00  !,f  ^^^"^  f  ^^^  ^^'  ^"  ''""'• 

The  sum  of  these  amounts, 

$31,836.  $31.83(5,  is  the  value  of  the 

p.  annuity  at  the  end  of  the  3d 

'  yr.    Or,  we  may  regard  the 

$10  X  1.06^        $10  _  13-j^  ggg^        annuity  as  the  1st  term  of 

1.06  —  1  a  Geometrical    Series  ;    the 

amount  of  %l  for  1  yr.,  at 
tJie  given  rate  as  the  ratio ;  and  the  number  of  years,  the  number  of 
terms,  and  perform  the  example  by  (527,  528). 


ANNUITIES 


407 


To  facilitate  computations,  the  following  table  answers  a 
very  good  purpose.  It  gives  the  amount  of  any  unit  for  any 
number  of  periods  from  1  to  40,  at  5,  6,  and  1%. 

TABLE 

Of  amounts  of  $1  or  £1  annuity  per  annum,  at  compound  interest. 


£ 

03 

1 

5  per  cent. 

6  per  cent. 

7  per  cent. 

1 

5  per  cent. 

6  per  cent. 

7  per  cent. 

1.000  000 

1.000  000 

1.000  000 

21 

35.719  252 

39.992  727 

44.865  177 

2 

2.050  000 

2.060  000 

2.070  000 

22 

38.505  214 

43.392  290 

49.005  739 

3 

3.152  500 

3.183  600 

3.214  900 

23 

41.430  475 

46.995  828 

53.436  141 

4 

4.310  125 

4.374  616 

4.439  943 

24 

44.501  999 

50.815  577 

58.176  671 

5 

5.525  631 

5.637  093 

5.750  739 

25 

47.727  099 

54.864  512 

&3.249  030 

6 

6.801  913 

6.975  319 

7.153  291 

26 

51.113  454 

59.156  383 

68.676  470 

7 

8.142  008 

8.393  833 

8.6;>i  021 

^ 

54.669  126 

63.705  766 

74.483  823 

8 

9.549  1C9 

9.897  468 

10.259  803 

28 

58.402  583 

68.528  112 

80.697  691 

9 

11.026  564 

11.491  316 

11.977  989 

29 

62.322  712 

73.639  798 

87.346  529 

10 

12.577  893 

13.180  795 

13.816  448 

30 

66.438  848 

79.058  186 

94.460  786 

11 

14.206  787 

14.971  643 

15.78;3  599 

31 

70.760  790 

84.801  677 

102.073  041 

12 

15.917  127 

16.869  941 

17.888  451 

39 

75.298  829 

90.889  778 

110.218  154 

13 

17.712  983 

18.882  138 

20 140  643 

a3 

80.063  771 

97343  165 

118.933  425 

14 

19  598  632 

21.015  066 

22.550  488 

34 

85.066  959 

104183  755 

128  258  765 

15 

21.578  564 

23.275  970 

25.129  022 

35 

90.320  307 

111.434  780 

138.236  878 

16 

23.657  492 

25.670  528 

27.838  054 

m 

95.836  323 

119.120  867 

148.913  460 

17 

25.840  366 

28.212  880 

30.840  217 

37 

101.628  139 

127.268  119 

160.337  400 

18 

28.132  385 

30.905  653 

a3.999  033 

38 

107.709  546 

135.904  206 

172.561  020 

19 

30.539  004 

33.759  992 

37.378  965 

39 

114.095  023 

145.058  458 

185.640  292 

20 

33.065  954 

36.785  591 

40.995  492 

40 

120.799  774 

154.761  966 

199.635  112 

To  use  this  table  we  have  the  following 

531.  EuLE. — Find  the  amount  of  1  unit  for  the  given 
time  and  rate,  and  multiply  this  amount  hy  the  number  of 
units. 

Example. — To  what  sum  will  an  annuity  of  S225  amount 

in  12  yr.,  at  7^  ? 

SOLUTION.  Explanation. — We  find 

$17.888451   X  225  =  $4024.901.       in  the  table,  under  the  head- 
ing 7%,  and  opposite  12  in 
the  left-hand  column,  the  number  17.888451,  which  is  the  number  of 


408  PROGEESSIONS. 

dollars  to  which  an  annuity  of  $1  will  amount  in  12  yr,,  at  7%.  Mul- 
tiplying this  by  the  given  number  of  dollars,  225,  we  have  $4024901, 
the  value  of  an  annuity  of  $225  running  12  yr.,  at  7%. 

PROBLEMS  . 

1.  A  gentleman  deposited  for  his  son  $150  for  10  consecu- 
tive years  in  a  savings  bank  that  paid  b%  compound  in- 
terest. To  what  sum  did  the  annuity  amount  immediately 
after  the  10th  payment  ?  Ans,  $1886.084. 

2.  A  gentleman  paid  $400  a  year  rent,  for  15  yr.  To  what 
sum  did  his  15  yr.  rent  amount,  at  7^  ?     Ans.  $10051.61. 

3.  A  merchant  at  the  age  of  30  insured  his  life  for  $10000, 
paying  a  premium  of  $23.30  on  $1000.  He  died  just  after 
making  the  30th  annual  payment.  How  much  more  money 
would  his  family  have  received  had  this  premium  been 
improved  at  6^  compound  interest  ?  Ans,  $8420.56. 

4  A  father  desired  to  deposit  annually  in  a  savings  bank 
such  a  sum  as  would  amount  in  10  payments,  at  5^,  to 
$1886.684.    What  must  be  the  annual  deposit  ? 

Solution. — Since  in  10  yr.,  at  5  % ,  $1  annuity  would  amount  to 
$12.5779  +  ,  it  will  require  as  many  dollars  annually  to  amount  to 
$1886.684,  as  $12.5779  +  is  contained  times  in  $1886.684,  or  150  times. 
Therefore,  he  must  deposit  annually  $150,  Ans. 

5.  A  dealer  in  real  estate  sells  $600  lots  for  $100  cash, 
and  the  balance  in  10  equal  annual  payments,  which  pay 
both  principal  and  interest,  at  7^.  How  much  is  each  pay- 
ment? 

Solution.— In  10  years,  at  7%,  compound  interest,  $500  will 
amount  to  $983.5755  (378).  An  annuity  of  $1  for  10  yr.,  at  7% ,  pro- 
duces $13.81645  (see  Table).  It  will,  therefore,  take  an  annuity,  or 
yearly  payment  of  as  many  dollars  as  $13.81645  is  contained  times  in 
$983.5755  to  yield  $983.5955  in  10  yr.  $983.5755  -^  $13.81645  =  71. 189. 
The  annual  payment  is,  therefore,  $71,189,  Ana. 

6.  What  would  have  been  the  annual  payments  at  &%  ? 

Ans.  $67,934. 


OUTLINE    OF    MENSURATION. 


t 

O 


TERMS. 


AREAS. 


OF 
LINES 


■  V. 


533.  ^  Surface. 

534.  An  Area. 

535.  A  Triangle. 

536.  Tlie  Base. 

537.  The  Altitude. 

538.  Tlie  Diagonal. 

539.  To  find  area  of  any  Parallelogram. 

540.  To  find  area  of  any  Triangle  {A). 

541.  To  find  area  of  any  Trapezoid. 

542.  To  find  area  of  any  Triangle  (li). 

543.  To  find  area  of  any  Plane  Figure. 

544.  To  find  area  of  any  Regular  Figure. 

545.  548.  To  find  area  of  any  Circle. 


546.  To  finil  Circumference. 
47,  To  find  Uiameter. 


549.  A  Solid. 

'550.  Convex  Surface. 
551.  Entire  Surface. 
SURFACE,    -i    552.  To  find  surface  of  Prism. 

553.  To  find  surface  of  Pyramid. 

554.  To  find  surface  of  Frustum. 


555.  A  Cylinder. 
557.  A  Cone. 
559.  A  Frustum. 
561.  ^  Sphere. 

563. 


656.  To  find  Surface. 
658.  To  find  Surface. 
660.  To  find  Surface. 
562.  To  find  Surface. 


VOLUME. 


40Q 


To  find  volume  of  Prism  or  Cylin- 
der. 

564.  To  find  volume  of  Pyramid  or  Cone, 

565.  To  find  volume  of  Frustum. 
.   566.   To  find  volume  of  Sphere. 


CHAPTER    IX 


A 


MEM  SJXIR^TiId^ 


532.  Mensuration  treats  of  the  measurement .  of 
magnitudes. 

In  reference  to  the  kinds  of  magnitude,  mensuration  is  of  four  kinds, 
namely,  Lines,  Angles,  Suj'faces,  and  Solids. 

A  Straight  Line  is  one  that  does  not  change  its  direction. 

Lines  and  Angles,  and  Surfaces  and  Solids  in  part,  have  been  some- 
what fully  discussed  in  Arts,  202-224,  and  also  247-258.  It 
is  proposed  in  this  Chapter  simply  to  complete  the  subject  of  mensura- 
tion, so  far  as  may  be  proper  in  a  work  of  this  kind,  by  adding  some 
Definitions,  Explanations,  and  Rules,  which  the  pupil  is  now  prepared 
to  comprehend,  with  reference  to  surfaces  and  solids. 

SQUABE   MEASUHB. 

533.  A  Surface  is  that  which  has  length  and  breadth 
without  thickness  (204). 

534.  An  Area  is  a  definite  amount  of  surface. 

535.  A  Triangle  is  a  plane  surface  bounded  by  three 
straight  lines.    (Fig.  1.) 

Triangles  are  equilateral,  isosceles,  and  scalene,  when  the  three  sides 
are  equal,  when  two  sides  are  equal,  and  when  no  two  sides  are  equal, 
respectively. 

They  are  also  right-angled,  obtuse-angled,  acute-angled,  and  equi- 
angular, when  they  have  one  right-angle,  when  they  have  OTie  obtuse- 
angle,  when  all  the  angles  are  acute,  and  when  all  the  angles  are  eqvM, 
respectively. 

41U 


MEKSURATIOiq- 


411 


Fig 


ABC,  Fig.  1,  is  a  right-angled  trian- 
gle, and  also  a  scalene  triangle. 

ADC,  Fig.  2,  and  D  C  B,  Fig.  4,  are 
obtuse-angled-triangles,  and  also  scalene  tri- 


FiG.  2. 


Fig.  3. 


Fig.  4. 


A  Quadrilateral  is  a  surface  bounded  by 
four  straight  lines,  Figs.  2,  3,  and  4. 

Quadrilaterals  are  subdivided  into  par- 
allelograms,   trapezoids,    and    trapeziums, 
that  is,  four-sided    figures  having   their 
opposite    sides  parallel,  ABEC,  Fig.  3; 
having  two  sides   parallel    and   two    in- 
clined to  each  other,  Fig.  4;  having 
no  two  sides  parallel.  Fig.  2,  respec- 
tively. 

Lines  are  said  to  be  parallel  when 
they  will  not  meet,  however  far  they 
may  be  produced,  as  C  E  and  A  B, 
Fig.,3 ;  and  D  C  and  A  B,  Fig.  4. 

Parallelograms  include  rectangu- 
lar and  square  fig- 
ures (207,  208); 
the  rhombus,  all  of 
whose  sides  are  equal 
and  angles  oblique ; 
and  the  rhomboid, 
whose  opposite  sides 
are  equal  and  angles 
oblique,  ABEC, 
Fig.  3. 

A  surface 
bounded  by 
five  sides  is 
called  a  pen- 
tagon ;  one 
B      by  six  sides. 


412  MENSURATION^. 

a  hexagon  ;  by  seten,  a  heptagon  ;  by  eight,  an  octagon  ;  by  ^«;i,  a  deca- 
gon ;  by  eleven,  an  undecagon;  by  twelve,  a  dodecagon. 

536.  The  JBase  of  a  plane  figure  is  the  side  on  which 
it  is  supposed  to  rest,  as  B  0,  Fig.  1 ;  A  B,  Figs.  2,  3,  and  4. 

537.  The  Altitude  is  the  perpendicular  distance  be- 
tween the  base  and  the  vertex  of  the  opposite  angle,  or 
between  the  base  and  the  opposite  side ;  as  A  B,  Fig.  1,  and 
C  D,  Fig.  3,  and  D  E,  Fig.  4. 

538.  The  Diagonal  of  a  plane  figure  is  the  straight 
line  joining  the  vertices  of  two  angles  not  connected  by  one 
of  the  sides;  as  A  0,  Fig.  2,  and  D  B,  Fig.  4. 

In  Fig.  3,  if  we  cut  off  the  part  ADC  and  place  it  to  the  right  of 
the  figure  C  D  B  E,  so  that  A  C  will  coincide  with  B  E,  we  see  that 
A  B  E  C  will  be  changed  to  a  rectangle  (207),  which  is  equivalent  to 
A  B  E  C,  and  whose  base  is  equal  to  A  B,  and  altitude  C  D  ;  and  we 
find  the  area  by  multiplying  together  the  two  dimensions  A  B  and 
C  D  (209).  Therefore,  to  find  the  area  of  any  parallelogram  we 
have  this 

539.  EuLE. — Multiply  the  base  hy  the  altitude. 

And  since  a  triangle  is  one-half  of  the  parallelogram  having  the 
same  base  and  altitude,  (See  ABC,  Fig.  3,)  we  find  its  area  by  this 

540.  EuLE.  —  A.  Multiply  the  lase  hy  one-half  the 
altitude. 

And  since  a  trapezoid  is  only  two  -triangles,  BCD  and  BAD,  Fig. 
4,  and  they  have  the  same  altitude,  D  E,  and  the  base  of  one  is  A  B, 
and  D  C  may  be  regarded  as  the  base  of  the  other,  we  may  take  the 
sum  of  the  areas  of  the  two  triangles  as  the  area  of  the  trapezoid. 
And  as  we  have  the  area  of  one  of  the  triangles  equal  to  one-half  its 
base  multiplied  by  its  altitude,  and  the  area  of  the  other  triangle  equal 
to  one-half  its  base  by  the  same  altitude,  we  must  have  the  area  of  the 
trapezoid  equal  to 

54:1.  One-half  the  su7n  of  its  two  sides,  mMiiMed  hy  its 
altitude. 


MEI^SURATION.  41S 

The  area  of  a  triangle  is  also  found  by  this 

542.  EuLE. — B.  From  the  half  sum  of  the  three  sides 
subtract  each  side  separately.  Theti  extract  the  square  root 
of  the  continued  product  of  these  remainders  and  the  half 
sum  of  the  sides.. 

The  area  of  any  plane  figure  may  be  found 

543.  By  dividing  it  i^ito  triangles,  calculating  the  area 
of  each  triangle  separately,  and  then  finding  the  sum  of 
these  areas. 

If  a  figure  is  regular,  that  is,  has  its  sides  and  angles  equal,  each  to 
each,  its  area  may  be  found 

544.  By  multiplying  the  distance  around  it  {the  peri- 
meter), hy  one-half  the  perpendicular  distance  from  its  center 
to  one  of  its  sides. 

If  a  regular  figure  has  its  sides  exceedingly  small  and  an  exceed- 
ingly large  number  of  them,  the  sides  taken  together  form  the  circum- 
ference (223)  of  a  circle,  (222),  and  the  perpendicular  distance  froip 
the  centre  to  one  of  its  sides  is  the  radius  (223) ;  and,  therefore,  we 
9ay  that 

545.  Hie  area  of  a  circle  equals  the  product  of  the  cir- 
cumference,  hy  one-half  the  radius. 

It  is  proven  in  Geometry,  that  the  ratio  of  the  circumference  of  a 
circle  to  its  diameter  is  3.14159,  nearly.  By  means  of  this  truth  we 
deduce  various  rules  for  determining  different  parts  of  a  circle  when 
other  parts  are  known. 

Thus,  since  the  circumference  is  3.14159  times  the  diameter. 
To  find  the  circumference  when  the  diameter  is  known, 

546.  Rule. — Multiply  the  diameter,  or  twice  the  radius, 
by  3.U159. 

To  find  the  diameter  when  the  circumference  is  known, 

547.  Rule. — 1.  Divide  the  circumference  by  SJJf.159, 


414  MENSURATION". 

Or,  2.  Multiply  the  circumference  hy.81831  (  =  j. 

From  the  foregoing  rules  we  may  readily  deduce  the  following 
additional  rules  for  finding  the  area  of  a  circle. 

548.    Multiply  the  square  of  the  diameter  hy  .786 J^, 
Multiply  the  square  of  the  radius  hy  3.1j^159. 
Multiply  the  square  of  the  circumference  hy  ,079577, 

PROBLEMS. 

Find  the  area  of  a  triangle  whose 

1.  Base  is  25  ft,  altitude  6.  Ans.  75  sq.  ft. 

2.  Base  is  100  rd.,  altitude  50.  Ans.  2500  sq.  id. 

3.  Base  is  16  yards,  altitude  16.  Ans.  128  sq.  yd. 

4.  Base  is  37^  ft.,  altitude  25.  Ans.  468|  sq.  ft. 

5.  Sides  are  6,  10,  and  12  rods.  Ans.  29.933  sq.  rd. 

6.  Sides  are  9,  15,  and  20  yd.  Ans,  63.277  sq.  yd. 

7.  Sides  are  13,  26,  and  35  ft.  Ans,  139.77  sq.  ft. 

Find  the  area  of  a  parallelogram  whose 
»    8.  Base  is  13,  altitude  20  ft.  Ans.  260  sq.  ft. 

9.  Base  is  250,  altitude  10  rd.  Ans.  2500  sq.  rd. 

.10.  Base  is  l^,  altitude  16  yd.  Ans.  200  sq.  yd. 

11.  Base  is  37J  in.,  altitude  6  ft.  A7is.  2700  sq.  in. 

Find  the  area  of  a  trapezoid  whose  bases 

12.  Are  6  and  10  ft.,  altitude  5  ft.  Ans.  40  sq.  ft. 

13.  Are  25  and  35  yd.,  altitude  9  ft.         Ans.  90  sq.  yd. 

14.  Are  25  and  35  ft.,  altitude  9  yd.        Ans.  90  sq.  yd. 

15.  Are  75  ft.  and  17  yd.,  altitude  10  ft.   Ans.  70  sq.  yd. 

Find  the  area  of  a  trapezuim  whose  diagonal 

16.  Is  21,  perpendiculars  3  and  5  ft.  A7is.  84  sq.  ft. 

17.  Is  36,  perpendiculars  21  and  30  ft.     Ans,  918  sq.  ft. 

18.  Is  31,  perpendiculars  27  and  16  yd.    Ans.  666|^  sq.  yd. 

19.  Is  15,  perpendiculars  8  and  10  rd.     Ans,  135  sq.  rd. 


MENSURATION^. 


415 


Find  the  area  of  a  regular  figure  of 

20.  Five  sides,  each  10  ft;  per.  6.88  ft.     Ans.  172  sq.  tt. 

21.  Six  sides,  each  10  ft. ;  per.  8. 66  ft.    Ans.  259.8  sq.  ft 

22.  Eight  sides,  each  20  ft;  per.  24.14  ft. 

A71S.  1931.2  sq.  ft. 
Find  the  circumference  of 

23.  A  circle  whose  diameter  is  10.  Ans.  31.4159. 

24.  A  circle  whose  diameter  is  25.  A7is.  78.54. 

25.  A  circle  whose  radius  is  10.  Ans.  6^.8318. 

26.  A  circle  whose  radius  is  100.  Ans.  628.318. 

Find  the  diameter  of 

27.  A  circle  whose  circumference  is  36.    Ans.  11.45916. 

28.  A  circle  whose  circumference  is  25.  Ans.  7.958 

Find  the  area  of  a  circle  whose 

29.  Circumf.  is  20  ft ;  radius,  3.1831.  Ans.  31.831. 

30.  Circumf.  is  25  ft ;  radius,  3.97887.      Ans.  49.7359. 


31.  Diameter  is  13  rd. 

32.  Diameter  is  16  yd. 

33.  Eadius  is  35  ft 

34.  Circumference  13.5  ft. 

35.  Circumference  12.5  rd. 


Ans.  132.7326  sq.  rd. 

Ans.  201.0624  sq.  yd. 

A71S.  3848.46  sq.  ft 

Ans.  14.5  sq.  ft 

Ans.  12.434  sq.  rods. 


SOLIDS. 

549.  A  Solid f  or  Body^  is  that 
which  has  length,  breadth  and  thick- 
ness, or  three  dimensions. 

A  JPrism  is  a  solid,  bounded  by  plane  sur- 
faces, two  of  which,  the  ends,  or  bases  of  the 
prism,  are  equal  and  similar  plane  figures ; 
and  the  sides,  or  faces,  parallelograms ;  as 
Figs.  1,  2,  and  3,  and  (215,  Fig.  3.) 


416 


MENSURATION. 


Fig  3. 


Fig.  4. 


Both  plane  fibres  and  solids  are  often  re- 
garded as  having  two  bases,  an  upper  and  a 
lower  base.  In  Fig.  1,  A  B  C  is  the  lower, 
and  D  E  F,  the  upper  base.  In  Fig.  2,  A  B  C  D, 
the  lower,  and  E  F  G  H  the  upper  base. 

Prisms  are  named  from  the  form  of  their 
bases.  Fig.  1  represents  a  triangular  prism, 
because  its  bases  ABC  and  D  E  F  are  tri- 
angles;  Fig.  2,  a  quadrangular  prism,  be- 
cause its  bases  A  B  C  D  and  E  F  G  H  are 
quadrilaterals;  Fig.  3,  a  pentagonal  prism, 
because  its  bases  ABODE  and  FGHIJ 
are  pentagons. 

The  edges  of  a  prism  are  the  lines  in  which 
the  bounding  surfaces  meet.  Thus,  in  Fig.  1, 
A  D,  C  B  and  B  E  are  edges ;  in  Fig.  2,  A  E, 
B  F,  &c.  are  edges. 

A  JRlght  Prism  is  one  whose  faces  are 
perpendicular  to  the  base.  Fig.  3  is  a  right 
prism,  because  each  of  its  faces,  as  A  B  G  F, 
is  perpendicular  to  the  base  A  B  C  D  E. 

A  Pyramid  is  a  solid  having  for  its  bases 
any  polygon,  and  for  the  rest  of  its  surface, 
plane  triangles.  Fig.  4.  It  terminates  in  a 
point  called  its  vertex,  S,  Fig.  4. 

Pyramids  are  named  from  their  bases.  The 
one  represented  in  the  margin  is  a  pentan- 
gular pyramid,  because  its  base  is  a  pentagon. 

The  altitude  of  a  pyramid  is  the  perpen- 
dicular distance  from  its  vertex  to  its  base. 
If  the  pyramid  is  a  rigM  pyramid,  the  perpen- 
dicular will  fall  on  the  middle  of  the  base ;  as, 
S  0,  Fig.  4. 

The  slant  height  is  the  length  of  a  line 
drawn  from  the  vertex  perpendicular  to  one 
side  of  the  base ;  as,  S  M,  Fig.  4. 


MENSURATION. 


417 


Fig.  5. 


The  frustum  of  a  pyramid  is  the 
part  that  remains  after  the  top  is 
cut  off  by  a  plane  parallel  to  the 
base,  Fig.  5.  A  B  C  D  E  repre- 
sents the  lower,  and  fghij  the 
upper  base.  They  are  named  from 
their  bases  ;  the  one  represented  in 
the  margin  being  pentagonal. 

The  altitude  of  a  frustum  is  the 
perpendicular  distance  between  its 
bases. 


The  slant  height  of  a  frustum  is  the  perpendicular  distance  be- 
tween the  two  parallel  sides  of  one  of  its  faces ;  as,  L  K,  Fig.  5. 


SURFACES   OF  SOLIDS, 

550.  The  Convex  Surface  of  a  prism,  or  pyramid, 
or  frustum,  is  all  of  its  surface  except  the  base,  or  bases. 

551.  The  Entire  Surface  is  the  convex  surface  and 
the  surface  of  the  base  or  bases. 

By  insi)ection  we  can  readily  see  that  in  order  to  obtain  the  surface 
of  solids  we  have  simply  to  apply  the  rules  already  learned.  For 
example — to  find  the  convex  surface  of  the  prism.  Fig.  1,  we  have 
only  to  find  the  area  of  the  three  parallelograms  A  B  E  D,  B  E  F  C,  and 
A  C  F  D,  and  then  find  their  sum,  which  is  just  the  same  as  multiply- 
ing the  sum  of  A  B,  B  C,  and  C  A  by  the  altitude  A  D.  If  we  wish 
the  entire  surface,  we  add  the  area  of  the  two  triangles,  ABC,  and 
D  E  F.     Hence,  to  find  the  surface  of  a  prism  we  have  this 

552.  Rule. — Multiply  the  perimeter  of  the  base  hy  the 
altitude.  To  this  result,  add  the  area  of  the  bases  for  the 
entire  surface. 

In  like  manner,  to  find  the  convex  surface  of  a  pyramid,  we  find  the 
sum  of  the  areas  of  the  triangles  that  form  its  surface ;  the  slant 
27 


418  MENSUEATIOis. 

height  of  the  pyramid  being  the  altitude  of  the  triangles.    From  which 
we  deduce  this 

553.  EuLE. — Multiply  the  perimeter  of  the  base  by  one- 
half  the  slant  height.  To  this  add  the  area  of  base,  for  the 
entire  surface. 

Again,  since  the  surface  of  the  frustum  of  a  pyramid  is  composed 
of  trapezoids,  we  see  that  the  surface  can  be  obtained  by  the  following : 

554.  EuLE. — Multiply  the  sum  of  the  perimeters  of  the 
parallel  bases,  by  one-half  the  slant  height.  Add  the  areas 
of  the  bases,  for  the  entire  surface. 


P  ROBIjEMS. 

What  is  the  convex  surface  of  a  prism  whose 

1.  Altitude  is  10  ft.  and  perimeter  of  base  16  ft.  ? 

Ans.  160  ft. 

2.  Altitude  is  13  yd.  and  perimeter  of  base  14  yd.  ? 

Ans.  182  sq.  yd. 

3.  Altitude  is  6J  in.  and  perimeter  of  base  14  in.  ? 

Ans.  87^  in. 

4.  What  is  the  entire  surface  of  a  square  prism  whose 
altitude  is  25  ft.  and  each  side  of  the  base  7  ft.  ? 

Ans.  798  sq.  ft. 

5.  What  is  the  entire  surface  of  a  triangular  prism  whose 
altitude  is  5  yd.  and  each  side  of  the  base,  6  yd.  ? 

Ans.  121.176  sq.  yd. 

6.  What  is  the  convex  surface  of  a  pentangular  pyramid 
whose  slant  height  is  30  ft.  and  each  side  of  its  base  is  3  ft.  ? 

Ans.  225  sq.  ft. 

7.  What  is  the  entire  surface  of  a  quadrangular  pyramid 
whose  slant  height  is  25  ft.  and  each  side  of  the  base  is  6  ft.  ? 

Ans.  336  sq.  ft. 


MENSURATIOif. 


419 


8.  What  is  the  entire  surface  of  a  triangular  pyramid 
whose  slant  height  is  35  in.  and  each  side  of  its  base  is  6  in.  ? 

Ans.  330.588  sq.  in. 

9.  What  is  the  convex  surface  of  the  frustum  of  an  octa- 
gonal pyramid  whose  slant  height  is  20  ft.  and  each  side  of 
whose  lower  base  is  6  ft.  and  upper  base  4  ft.  ? 


Fig.  6. 


Ans.  800  sq.  ft. 

555.  A  Cylinder  is  a  solid  whose 

ends  are  equal  parallel  circles,  Fig.  6. 

When  the  line 'that  joins  the  centers  of  the 
bases  is  perpendicular  to  the  bases,  the  cylinder 
is  called  a  Right  Cylinder ;  the  line,  MN, 
Fig.  6,  is  the  axis  and  also  the  altitude  of  the  cyl- 
inder ;  and  since  the  cylinder  is  a  prism  with  a 
great  number  of  faces,  to  find  its  convex  and 
entire  surface,  we  have  only  to  write  "  circum- 
ference "  for  'perimeter  in  the  rule  for  finding  the  surface  of  a  prism, 
and  obtain  this 

556.  Rule. — Multiply  the  circumference  of  the  base  ly 
the  altiticde.  To  this  result  add  the  areas  of  the  bases,  for 
the  entire  surface. 

5511..  A  Cone  is  a  solid  having  a 
circle  for  its  base  and  tapering  to  a 
point  called  its  vertex,  S,  Fig.  7. 

The  altitude  is  the  perpendicular  distance 
from  the  vertex  to  the  base.  The  slant  height 
is  the  shortest  distance  from  the  vertex  to  the 
circumference  of  the  base,  AS,  Fig.  7.  A  cone  is 
a  pyramid  with  a  great  number  of  faces ;  and 
hence  to  find  its  surface  we  use  the 

55S.  Rule. — Multiply  the  circum- 
ference of  the  base  by  one-half  the  slant  height.  To  this 
add  the  area  of  base,  for  the  entire  surface. 


420 


MEKSURATIOK 


Fig.  8.  559.  The  JFrustufii  of  a  Cone 

is  the  part  left  after  the  top  has  been 
cut  off  by  a  plane  parallel  to  the  base, 
Fig.  8. 

The  artist  has  done  his  work  so  well  that  we 
need  not  define  altitude  (line  representing  the 
axis),  slant  height  and  radius.  To  find  the 
surface  we  have  the 

560t  EuLE. — Multiply  the  sum  of  the  circumferences  of 
the  two  hases  hy  half  the  slant  height,  and  add  the  areas  of 
hoses  for  the  entire  surface. 

Fig.  9.  561.  A  Sphere  is  a  volume  bounded 

by  a  uniformly  curved  surface,  every 
point  of  which  is  equally  distant  from  a 
point  within  called  the  center,  Fig.  9. 


The  Diameter  of  a  sphere  is  a  straight  line 
passing  through  the  center  and  terminating  at 
both  ends  with  the  surface,  MN,  Fig.  9.  The 
Radius  is  the  distance  from  the  center  to  the  surface,  OS,  Fig.  9. 
The  Circumference  is  the  greatest  distance  around  the  sphere. 

The  relations  of  a  diameter,  radius,  and  circumference  of  a  sphere 
are  the  same  as  those  that  the  diameter,  radium,  and  circumference  of 
a  circle  bear  to  each  other. 

To  find  the  surface  of  a  sphere  we  use  this 

563.   EuLE. — Multiply  the  diameter  hy  the  circumfer- 
ence.   Or, 
Multiply  the  square  of  the  diameter  hy  S.lJflG, 


P110B1.EMS. 

Find  the  convex  surface  of  a 
1.  Cylinder,  alt.  10  ft. ;  circum.  of  base  31.83. 

Ans.  318.3  sq.  ft 


MENSURATION".  421 

2.  Cylinder,  alt.  25  ft. ;  radius  of  base  10  ft. 

Ans.  1570.8  sq.  ft. 

3.  Cone,  slant  height  20  ft. ;  radius  of  base  5  ft, 

Ans.  314.16  sq.  ft. 

4.  Frustum  of  a  cone  whose  slant  height  is  36  ft. ;  radius 
of  upper  base  10  ft.,  of  lower  base  15  ft.     Ans.  2827.44  sq.  ft 

5.  Find  entire  surface  of  frustum  of  a  cone,  slant  height 
80  ft. ;  radius  of  upper  base  5  ft.,  and  lower  base  2^  ft. 

Ans.  569.415  sq.  ft. 

6.  Find  the  surface  of  a  sphere  whose  diameter  is  25. 

A71S.  25  X  3.1416  X  25  r=  1963.5. 

7.  Find  the  surface  of  a  sphere  whose  radius  is  7  rd. 

Ans.  615.7536  sq.  rd. 

VOLUMES. 

To  find  the  volume  of  a  prism  or  cylinder, 

563.  Multiply  the  area  of  the  base  by  the  altitude. 
To  find  the  volume  of  a  pyramid  or  cone, 

564.  Multiply  the  area  of  the  base  by  ^  the  altitude. 
To  find  the  volume  of  the  frusttim  of  a  pyramid  or  cone, 

5G5.     To  the  sum  of  the  areas  of  the  two  bases,  add  the 

square  root  of  their  jjroduct,  and  multiply  this  result  by  ^  of 

the  altitude. 

To  find  the  volume  of  a  sphere, 

5GG,     Mtiltiply  the  cube  of  the  diameter  by  .5236, 

Or,  Multiply  the  cube  of  the  radius  by  Jf.l888. 

PR  on  LJEM  S. 

1.  Find  the  capacity  of  a  cylindrical  measure  18-|-  inches 
in  diameter,  and  8  in.  deep.  Ans.  1  bu. 

2.  Of  a  cylindrical  bucket  5  inches  in  diameter  and  7  in. 
deep  (wine  measure).  Ans.  2.38  qt. 


433  MENSUKATIOK. 

3.  Find  the  volume  of  a  triangular  bar  15  ft.  long,  and 
whose  sides  are  1  in.,  1  in.,  and  1.414  in.     Ans.  90  cu.  in. 

4.  How  many  conical  glasses,  each  3  inches  in  diameter 
and  4.902  in.  deep,  can  be  filled  from  a  qt.  bottle?  Ans.  5. 

5.  What  is  the  volume  of  a  quadrangular  pyramid,  eacli 
side  of  the  base  being  100  ft.  and  the  altitude  75  ft.  ? 

Ans.  250000  cu.  ft. 

6.  What  is  the  volume  of  the  frustum  of  a  pyramid  24  ft. 
high,  6  ft.  square  at  one  end  and  4  ft.  square  at  the  other  ? 

Ans.  608  cu.  ft. 

7.  How  many  bbl.  will  a  cistern  12  ft.  deep,  Avith  upper 
diameter  9  ft.  and  lower  diameter  3  ft.,  hold  ?   Ans.  87.3  bbl. 

8.  How  many  cubic  feet  of  iron  in  a  cannon  ball  36  in. 
in  diameter?  Ans.  14.1372. 

9.  How  many  cubic  ft.  of  granite  in  a  cylindrical  monu- 
ment 36  ft.  high,  with  a  base  whose  radius  is  2  ft.  ? 

10.  How  many  square  feet  of  sheet-iron  J  in.  thick  can 
be  made  of  a  cylindrical  shaft  20  ft.  long  and  4  inches  in 
diameter  ?  Ans.  83.776  sq.  ft. 

11.  What  are  the  relative  values  of  a  ball  of  gold  4  inches 
in  diameter  and  a  cylinder  of  same  diameter  and  altitude  ? 

Ans.  2  :  3. 

12.  What  is  the  capacity  of  an  oval  lime-kiln,  whose  top, 
bottom,  and  greatest  diameters  are  6,  2,  and  7  ft. ;  the  dis- 
tances of  the  top  and  bottom  diameters  from  the  greatest 

diameter  being  5  and  15  ft.  ?  A7is.  370.26  bu. 

« 

Multiply  the  sum  of  the  square  of  the  top  diameter  and  twice  the 
square  of  the  greatest  diameter  by  the  distance  between  these  diam- 
eters. Do  the  same  with  the  bottom  and  greatest  diameter  and  their 
distance.     Then  multiply  the  sum  of  these  two  results  by  .2G18. 

[(36  +  2  X  49)  X  5  +  (4  +  3  X  49)  x  15]  x  .2618  =  575.96  cu.  ft. 


THE     METRIC     SYSTEM.  423 

THE   METRIC   SYSTE3I. 

567.  The  Metric  System  is  a  decimal  system  of 
Weights  and  Measures^  founded  on  a  certain  unit  of  length 
called  the  metre  =  39.37079  inches. 

The  system  is  an  exceedingly  valuable  one,  bat  its  limited  use  in 
this  country  does  not  justify  our  treating  the  subject  as  fuUy  as  its 
importance  deserves. 

The  Metric  System  adopts  a  standard  unit  of  measure, 
and  then  forms  lower  denominations  by  the  decimal 
snh-measures  of  this  unit;  and  higher  denomina~ 
tions  by  the  decimal  multiples.  Thus,  taking  the  metre  as 
the  unit,  and  using  the  Latin  prefixes,  deci  (tenth),  centi 
(hundredth),  milli  (thousandth),  we  have 

Deci-mQtYQ  (dess'-e-meet-ur)  =.1  metre  (meet'ur). 

Centi-TCLQivQ  (cent'-e-meet-ur)  =  .01  metre. 

Milli-m.QtrQ  (mil'-e-meet-ur)  =  .001  metre. 

And  by  using  the  Greek  prefixes  deca  (ten),  lieda  (hun- 
dred), kilo  (thousand),  myria  (ten-thousand),  we  have 

Deca-TCieivQ  (dek'-a-meet-ur)  =  10  metres. 

Ifecta-metre  (hec'-ta-meet-ur)  =  100  metres. 

/u7o-metre  (kil'-o-meet-ur)  =  1000  metres. 

Myria-metre  (mir'-ee-a-meet-ur)  =  10000  metres. 

In  like  manner  taking  the  litre  as  the  unit  of  meas- 
urCf  we  have  milU-litre,  centi-litTe,  deci-litTe,  deca-litre,  &c. 
The  Units  of  Measure  are 

568.  The  Metre ,  the  Linear  Unit,  nearly  1  ten-mil- 
lionth of  the  distance  from  the  equator  to  either  pole. 

569.  The  Are  (air),  which  is  the  Unit  of  Surface  Meas- 
ure, and  is  a  square  whose  side  is  equal  to  10  metres,  or  1 
decametre. 

570.  The  Stere  (stair),  which  is  the  Unit  of  Solid 
Measure,  and  is  a  cube  whose  edge  is  1  metre. 


424  THE     METRIC     SYSTE 


571.  The  Litre  (leet-ur),  which  is  the  Unit  of  Measures  of 
Capacity,  and  is  a  cube  whose  edge  is  .1  metre,  or  1  decimetre. 

573.  Gratnme  (gram),  which  is  the  Unit  of  Weight, 
This  is  a  cube  of  pure  water  whose  edge  is  .01  metre,  or 
1  centimetre,  weighed  in  a  vacuum,  at  a  temperature,  when 
its  density  is  greatest,  39.2°  Fahrenheit's  Thermometer. 

JsrUMBBATION  AND   NOTATION. 

573.  Each  standard  unit  is  represented  by  the  initial  letter  of  its 
name.  Thus,  M,  for  metre  ;  A,  for  are  ;  S,  for  stere,  &c.  A  multiple  is 
represented  by  the  first  letter  of  its  prefix  followed  by  the  first  letter 
of  the  standard  unit ;  and  a  sub-multiple  by  the  first  letter  of  its  prefix 
followed  by  the  initial  letter  of  the  standard  unit  written  in  small  letters. 
Thus,  MM.  stands  ioi  myriametres,  mm.  for  millimetres  ;  DM.  stands  for 
decametres,  dm.  for  decimetres,  &c.,  &c.,  though  in  general,  since  all  de- 
nominations are  decimal,  the  name  of  the  standard  unit  is  the  only  name 
that  is  necessary  to  express  a  number.  Thus,  21.632  M.,  or  M.  21.632, 
OT2\y^Q^2\^iG2i.^  twenty-one  and  six  hundred  thirty-two  thousandths 
metres  ;  32.3  L.,  or  L.  32.3  is  read  thirty-two  and  three-tenths  litres, 

TABLES. 
Long  and  Linear  Measure. 

.1   I   I   I   i    1    I  I 

31476.3     4     1 
Read,  thirty-one  thousand  four  hun- 
dred  seventy-six   and    three   hundred 
forty-one  thousandths  metres. 

Note.  —  Observe    the    resemblance 
between  M.  31476.341,  and  $31476.341. 

Instead  of  placing  the  separatrix  between  the  metres  and  decimetres, 
which  we  do  in  measuring  cloth  and  ordinary  distances,  we  may  place 
it  between  the  kilometres  and  hectometres,  which  we  do  when  measur- 
ing long  distances,  as  the  length  of  rivers,  &c.  The  number  should 
then  be  read  31  and  476341  hundred-thousandths  kllotnetres. 

By  omitting  the  separatrix,  the  number  maybe  read  31476341  mil' 
Ihnetres. 


10  mm. 

1  Centimetre. 

10  cm. 

1  Decimetre. 

10  dc. 

1  Metre. 

10  M. 

1  Decametre. 

10  DM. 

1  Hectometre. 

10  HM. 

1  Kilometre. 

10  KM. 

1  Myriametre. 

THE     METRIC     SYSTEM. 


425 


Surface  or  Square  Measure. 

This  measure  is  used  in  measuring  land  and  other  sur- 
faces. The  Unit  of  Measure  is  an  are  =  100  sq.  M.  = 
119.60  sq.  yd.,  or  a  hectare  =  2.471  A. 

i .  I 

■.A.  I  2  34.42 

HA.  :      Two  hundred  thirty-four 
and   forty-two    hundredths 
ares. 
The  denomination  centare  is  sometimes  written  centim'6. 

It  is  customary  also  to  express  the  area  of  ordinary  sur- 
faces by  the  square  of  the  linear  metre,  called  the  square 
meter  =  1550  sq.  in. 


TABLE, 

100  cen tares  =  1  are 

100  ares        =  1  hectare  . 


TABLE. 

100  sq.  cm.   =  1  sq,  dm. 
100  sq.  dm.  =  1  sq.  m. 


«5    B   a 
^  ^   « 

347.3017  is  read  three  hundred 
forty-seven  and  two  thousand  sevens- 
teen  ten-thousandths  square  me- 
tres. 


Solid  or  Cubic  Measures. 

The  denominations  decastere  and  dedstere  are  sometimes 
used  when  applied  to  the  measure  of  the  volunie  of  fire- 
wood and  building-timber,  but  ordinarily  the  stere  = 
35.316  cu.  ft. 


TABLE. 

10  decistere  =  1  stere. 
10  stere         =  1  decastere. 


37489.004  is  read  37489  and  4 
thousandths  steres. 


Volumes,  such  as  excavations,  embankments,  &c.,  are  ex- 
pressed by  the  cube  of  some  denomination  of  linear  meas- 
ure; thus. 


426 


THE     METRIC     SYSTEM 


TABLE. 

1000  cu.  millim.  =  1  cu.  centim. 
1000  cu.  centim.  =  1  cu.  decim. 
1000  cu.  decim.  =  1  cu.  metre  = 


1  stere. 


£      J      I     I 

^    ^    J  s 

15.113601603 

is  read  15  and 
113001003  hUllonths 
cubic  metres. 


Measure  of  Capacity, 

This  corresponds  to  our  Liquid  and  Dry  Measures.     The 
Litre  =  .908  qt.  Dry  Measure,  or  1.0567  qt.  Liquid  Meas- 
ure  (^1  kilogramme,  or  2.2046  lb.  Av.),  is  the  Unit  in 
Liquid  Measure,  and  the  Hectolitre  in  Dry  Measure. 
TABLE, 


10  millilitres 
10  centilitres 
10  decilitres 
10  litres 
10  decalitres 
10  hectolitres 


=  1  centilitre. 
=  1  decilitre. 
=  1  Litre. 

z=  1  decalitre. 
=  1  Hectolitre. 
=  1  kilolitre. 


3  2  1  .  4  7  5  6  is  read  321 
and  4756  ten-thousandths  Li- 
tres =  3.214.756  Hectolitres. 


Weight. 

This  corresponds  to  our  Troy,  Apoth 
dupois  Weights.  The  IJiiit  of  Measure 
=  15.432  grains. 

TABLE. 
10  Milligrammes  =1  Centigramme. 
10  Centigrammes  =1  Decigramme. 
10  Decigrammes  =1  Gramme. 
10  Grammes  =1  Decagramme. 

10  Decagrammes  =1  Hectogramme. 
10  Hectogrammes =1  Kilogramme, 
10  Kilogrammes    =1  Myriagramme. 
10  Myriagrammes=l  Quintal, 
10  Quintals  =1  Tonneau. 


ecaries,  and  Avoir- 
is  the  Gra^nme 


^  g.  s'S  :S  <S  CD  ^  ^  3 

3  3  5  17  8.5.3  56 
is  read  325178  and 
5256  ten-thousandths 
grammes  =  335- 
.1785256  K.  G.  = 
3.351785356   Quintals, 

uCC.,  &C. 


THE     METRIC     SYSTEM.  427 

While  the  gramme  is  the  unit  of  measure  in  weighing  gold,  silver, 
medicines,  &c  ,  the  kilogramme,  or  kilo,  as  it  is  often  called,  is  used  as 
the  unit  in  weighing  meat,  butter,  grain,  groceries,  &c.,  &c.  ;  and  the 
onneau  in  weighing  such  articles  as  coal,  hay,  &c. 

nBDvcTiojsr. 

Since  the  numbers  of  the  Metric  System  are  either  pure 
or  mixed  decimals,  one  denomination  may  be  reduced  to 
another  by  annexing  ciphers,  or  removi7ig  the  decimal  point 
to  the  right,  in  Descending  Reduction  ;  and  in  As- 
cending Reduction  by  pointing  off  from  the  right,  or 
removing  the  decimal  point  to  the  left,  and  prefixing  ciphers, 
when  necessary. 

Example  1. — Reduce  4  K.  M.  to  M. 

Solution. — Since  1  kilometre  is  1000  metres,  4  kilometres  are  4 
times  1000  metres,  or  4000  metres.  Or  4.  with  three  ciphers  annexed 
=  4000,  required  number  of  metres. 

Example  2. — Reduce  5275  mm.  to  M. 

Solution. — Since  1000  millimetres  are  1  metre,  5275  millimetres 
are  as  many  metres  as  5275  contains  times  1000,  or  5.275  metres. 

l^ROBLEMS, 

1.  Reduce  125  M.  to  cm. 

2.  Reduce  12500  cm.  to  M. 

3.  Reduce  81.875  HA  to  A. 

4.  Reduce  8187.5  A.  to  HA. 

5.  Reduce  4.703  sq.  KM.  to  M. 

6.  Reduce  4703000  sq.  M.  to  KM. 

7.  Reduce  45.625  S.  to  ds. 

8.  Reduce  456.25  ds.  to  S. 

9.  Reduce  6  cu.  M.  to  cm. 

10.  Reduce  6000000  cu.  cm.  to  M. 


428  LUMBERMEN'S     RULES 


ADDITIOIV  ANiy    SUBTBACTION. 

EuLE. — Reduce,  if  necessary,  the  iiumhers  to  the  same  de- 
nomination^  and  proceed  as  in  Addition  and  Subtraction  of 
Decimals  (126,  128). 

LUMBBB31EN'8   MULUS. 
574.     1.  For  finding  the  approximate  number  op  square 

FEET  OF  INCH  BOARDS  THAT   A  GIVEN   LOG  WILL  YIELD. — From  the 

diameter  of  the  smaller  end  in  inches  subtract  4,  and  multiply  the  re- 
mainder by  itself.  The  product  will  be  nearly  the  number  of  square  feet 
of  inch  boards  in  a  log  16  feet  long.  For  a  log  of  any  other  length,  mul- 
tiply this  result  by  as  many  sixteenths  as  there  are  feet  in  its  length. 

Example. — How  many  square  feet  of  inch  boards  in  a  log 
whose  least  diameter  is  40  in.  and  length  16  ft.  ?     36  ft.  ? 

Solutions.    (40-4)  x  (40-4)  =  1296,  Ans. 

(40-4)  X  (40-4)  X  ff  =  2916,  Ans. 

5115,  2.  For  finding  the  number  of  cubic  feet  of  square 
timber  in  a  given  log. — Multiply  the  length  of  fhe  log  in  feet  by  the 
square  of  the  mean  diameter  in  inches,  and  divide  this  product  by  324. 

The  mean  diameter  is  usually  the  ^  sum  of  the  ena  diameters. 

^^  In  order  to  obtain  the  same  results  as  those  published  in  some 
tables  now  in  use,  it  is  necessary,  before  applying  the  rule,  to  call  a 
diameter  of  13  in.,  13i  in. ;  20  in.,  20.3  in. ;  22,  22^;  25,  251 ;  26,  26^; 
28, 28.6  ;  31,  29f  ;  34,  33| ;  35,  35]^ ;  37,  37^ ;  38,  38^  ;  39,  39f  ;  40, 41 J ; 
41,  42f ;  42,  43^,  &c. 

Example. — How  many  cu.  ft.  of  hewn  timber  in  a  log 

42  ft.  long  and  36  inches  in  diameter?    In  a  log  38  ft.  long 

and  33  inches  in  diameter  ? 

42x362       ...         .^       . 
Solutions,     —hkj-  =  168  cu.  ft.,  Ans. 
324 

§?-^^?!  =  127.72  cu.  ft.,  Am, 

Qa'x 


MISCELLAi^EOUS     PROBLEMS.  429 


Miscellaneous   Prohlems, 

1.  Divide  .001  by  .000001,  and  multiply  the  quotient  by 
.0002.  Ans.  .2. 

2.  Find  the  width  of  the  narrowest  street  across  which 
stepping-stones  either  4,  5,  or  8-  ft.  long  will  exactly  reach. 

Ans.  40  ft. 

3.  The  Main  Centennial  building  in  Philadelphia  is  1880 
ft.  long  by  464  ft.  wide.  At  $200000  per  A.,  what  would  be 
its  cost?  Ans.  $4005142.33. 

4.  From  ^  hhd.  take  .90625  gal.  Ans.  32  gal. 

5.  M  and  N  receive  the  same  salary.  M  saves  \  of  his ; 
N  spends  1210  more  a  year  than  M,  and  at  the  end  of  7  yr. 
finds  himself  $70  in  debt.    What  is  their  salary  ? 

Ans.  $800. 

6.  If  100  men,  by  working  6  hr.  each  da.,  can  in  27  da.  dig 
18  cellars  each  40  ft.  long,  36  ft.  wide,  and  12  ft.  deep,  how 
many  ft.  long  may  256  cellars  be  to  require  240  men  81  da. 
of  8  hr.  each,  to  dig  27  ft.  wide,  and  18  ft.  deep  ? 

Ans.  24  ft. 

7.  Bought  a  hhd.  of  molasses  for  50^  a  gal.  How  must  I 
sell  it  per  gal.  so  as  to  make  20^  on  the  cost  of  the  whole, 
18  gal.  having  leaked  out  ?  Ans.  84^. 

8.  What  is  the  amount  of  $6.06  for  6  yr.  6  mo.  6  da., 
at  Q%  ?  Ans.  $8.43. 

9.  A  man  dying  leaves  his  property  to  be  divided  as  fol- 
lows :  40;^  to  his  wife ;  25^  of  the  remainder  equally  between 
his  two  daughters ;  and  the  remainder  among  his  4  sons  in 
proportion  to  their  ages,  which  are  respectively  10, 14,  16, 
and  20  yr.     What  %  of  the  property  does  each  son  receive  ? 

Ans.  Hifo ;  lOJ^  ;  12^ ;  and  15^. 


430  MISCELLANEOUS     PROBLEMS. 

10.  What  is  the  difference  between  the  true  and  bank  dis- 
count of  $1000  for  4  mo.,  at  8%  ?  (Days  of  grace  in  both 
cases.),  Ans.  $.727. 

11.  What  is  the  value  of  a  30  da.  bill  of  exchange  on 
London  for  £1000,  at  i%  prem.,  int.  off  at  7%  ? 

Ans.  $4859.60. 

12.  Wishing  to  raise  $2288.40  at  a  bank  which  charges 
1^%  a  month,  I  desire  to  know  how  large  a  note  I  must 
make  for  90  da.  in  order  to  raise  this  amount. 

Ans.  $2400. 

13.  Eequired  the  duty  at  20^  on  8780  lb.  raisins  invoiced 
at  9j^  per  lb.  Ans.  $158.04. 

14.  A  farm  was  sold  for  $8100,  which  was  S%  more  than 
its  value  :  what  would  have  been  the  gain  %  if  it  had  been 
sold  for  $8500  ?  Ans.  13-^^. 

15.  Clarke  invested  $63750  in  stocks  which  yielded  him 
the  1st  yr.  5%;  the  2d,  6%;  and  the  3d,  7^.  He  found 
that  his  income  for  the  3  yr.  was  just  $25  less  than  the 
cost  of  92  shares  of  stock.     How  many  shares  has  he  ? 

Ans.  510. 

16.  If  it  costs  $328  to  put  a  fence  around  a  rectangular 
field  50  rd.  long  and  32  rd.  wide,  how  much  less  will  it  cost 
to  inclose  a  square  field  of  equal  area  with  the  same  kind  of 
fence  ?  Ans.  $8. 

17.  What  is  the  length  of  a  side  of  the  greatest  square  that 
can  be  cut  out  of  a  circular  piece  of  card-board  36  in.  in 
diameter  ?  Ans.  25.4558  in. 

18.  The  parallel  sides  of  a  trapezoid  are  8  and  23.8 
rd.,  the  oblique  sides  124-  and  13  rd.,  and  the  altitude 
10  rd.  What  is  the  area  of  the  trapezoid  and  what  the 
length  of  the  diagonals? 

Ans.  Area,  159  P. ;  diagonals,  18.44  and  19.12 +  . 


MISCELLANEOUS     PROBLEMS.  431 

19.  A  gentleman  owning  40  A.  of  land,  in  the  form  of  a 
square,  made  a  circular  race-course  1  rd.  wide  and  as  long  as 
it  was  possible  to  make  it.  How  much  land  was  inside  the 
course  ?  How  much  in  each  corner  of  the  field  ?  And 
how  much  did  the  "  course  "  occupy  ? 

^^5.  29.8648  A. ;  2.146  A.;  1.5512  A. 

20.  How  many  perches  of  masonry  in  a  circular  stack 
120  ft.  high  and  15  ft.  diameter  at  bottom  and  6  ft.  at 
the  top;  and  having  a  circular  opening  6  ft.  in  diameter 
at  the  bottom  and  2  ft.  at  the  top? 

Ans.  379.53+  perches. 

21.  How  many  times  as  strong  is  a  joist  2 J  in.  wide  and 
12  in.  deep,  as  one  3^  in.  wide  and  9  in.  deep  ? 

Ans.  H  times. 

The  relative  strength  of  timbers  is  estimated  by  multiplying  the 
breadth  by  the  square  of  the  depth. 

22.  How  many  times  as  strong  is  a  joist  15  in.  deep  and 
2^  in.  thick,  when  supported  on  its  narrow  side  as  when 
supported  on  its  broad  side  ?  Ans.  Q  times. 

23.  How  many  cu.  in.  of  heated  air  in  a  soap-bubble  5  in. 
in  diameter  ?  Ans.  65.45  cu.  in. 

24.  When  gold  is  worth  $16  an  ounce,  T.,  what  is  the 
value  of  a  ball  of  gold  6  in.  in  diameter  ?   Ans.  $18380.26. 

A  cu.  in.  of  gold  weighs  11.144  oz.,  Av.,  or  10.1573  oz.,  T. 

25.  What  is  the  diameter  of  a  spherical  cannon-ball  that 
can  be  made  of  3560  lb.  of  iron  ?  Ans.  29.8526  +  in. 

A  cu,  ft.  of  cast  iron  weighs  441.62  lb. 

26.  How  many  half -inch  spherical  musket-balls  can  be 
made  of  25  lb.  of  lead  ?  Ans.  922. 66  -f- . 

A  cu.  ft.  of  lead  weighs  715.37  lb. 


432  MISCELLAI^EOUS     PROBLEMS. 

27.  A  carter  killed  his  horse  by  compelling  him  to  draw 
at  a  load  a  cart-bed  4  x  6  x  2  ft.,  full  of  gravel.  How  much 
of  a  load  was  this  ?  Ans.  2.88  Tons. 

A  cu.  ft.  of  gravel  weighs  120  lb. 

28.  A  stick  of  English  oak  1  in.  square  will  support  10000 
lb.  How  many  lb.  will  a  cylinder  of  2  in.  diameter  of  the 
same  wood  support  ?  Ans.  31415.9  lb. 

29.  A  1-inch  square  stick  of  poplar  will  support  a  weight 
of  7200  lb.  What  sized  cylindrical  stick  will  support  a 
weight  of  1  Ton?  Ans.  Diameter,  .5947+  in. 

30.  What  is  the  capacity  of  a  cistern  whose  upper  diam- 
eter is  15  ft.,  lower  diameter  8  ft.,  and  depth  20  ft.  ? 

31.  How  many  circular  openings  ^  of  an  inch  in  diameter 
will  let  out  from  a  shower  bath  the  water  running  in  through 
a  pipe  of  }  inch  bore  ?  Ans.  225. 

32.  How  many  bushels  of  rye  can  be  stored  in  a  bin  3 
meters  long,  2  high  and  2  wide  ?  Ans.  340.5  bu. 

33.  A  merchant  bought  31.™75  of  broadcloth  at  $3.50  per 
metre :  for  what  price  per  yard  must  he  sell  it  to  gain  20^  ? 

34.  The  distance  on  P.  R.  R.  from  Pittsburgh  to  New 
York  is  444  miles.  What  would  the  fare  amount  to  at 
$.01749  per  kilometre  ?  Ans.  $12.50. 

35.  When  gold  is  worth  112 J,  what  must  be  paid  in  green- 
backs for  a  letter  of  credit  for  £350,  premium  %%,  and  inter- 
est off  for  63  days  at  10^? 

36.  Sold  my  bank-stock  which  paid  regular  dividends  of 
10^,  at  an  advance  of  30^,  and  invested  in  Government 
10-40's  at  110.  Did  I  increase  or  diminish  my  income? 
and  if  so,  how  many  %  ? 

37.  How  many  pounds  of  sheet  zinc,  weighing  3  lbs.  to 
the  sq.  ft.  will  Une  the  bottom  and  sides  of  a  tank  8  x  10  ft. 
and  2i  ft.  high  ?  Ans.  510  lbs. 


MISCELLAl^EOUS      PEOBLEMS.  433 

38.  A  cistern  is  f  full  of  water,  and  after  48  gal.  are 
drawn  out,  it  is  j\  full :  liow  many  gallons  will  it  contain  ? 

A71S.  153|-  gal. 

39.  If  a  watch  which  gains  CJ  min.  in  8  hours,  is  set 
right  at  12  M.  on  Sunday  :  what  will  be  the  time  by  it  on 
the  following  Wednesday  at  3  P.  M.  ? 

A71S.  58  min.  35|  sec.  past  3. 

40.  If  by  working  12J  hours  a  day,  a  man  can  do  a  piece 
of  work  in  6 J  days,  how  long  will  it  take  him  to  do  it  if  he 
works  8J  hours  a  day  ?  Ans.  8|f  days. 

41.  A  man  engaged  to  do  a  piece  of  work  in  45  days. 
After  working  30  days  he  found  that  he  had  completed  just 
1^  of  it :  he  then  hired  his  "neighbor"  to  assist  him,  and 
together  they  finished  the  remaining  ^  in  10  days  ;  how 
long  would  it  have  taken  the  "neighbor"  to  do  the  whole 
job  ?  Ans.  30  days. 

42.  f  of  William's  money  -f-  |  of  Samuel's  money  having 
been  placed  at  interest  for  6  yrs.  6  mos.,  S%,  yielded  $780. 
How  much  money  has  each,  if  |^  of  William's  money  equals 
J  of  Samuel's  ?     Ans.  William  has  $900 ;  Sam.  has  $1200. 

43.  The  distance  from  A  to  B  is  804  miles.  If  C  leaves 
A  at  the  same  time  that  D  leaves  B,  and  they  travel  toward 
each  other,  how  many  hours  will  they  be  in  meeting,  if  C 
travels  6-J  miles  and  D  5|  miles  an  hour?       Ans.  HO-^ifj. 

44.  My  house  cost  |  as  much  as  my  farm,  and  my  farm 
four  times  as  much  as  my  stock.  What  was  the  entire  cost 
of  my  house,  farm,  and  stock,  if  my  house  cost  $3000? 

Ans.  $8625. 

45.  After  spending  f  of  my  income  for  boarding  and  one- 
^  half  of  the  remainder  for  clothing,  I  find  that  I  have  saved 

$300  out  of  my  annual  income.     What  was  my  income  ? 

Ans.  $2400. 


434  MISCELLANEOUS      PR0BLE5IS. 

46.  By  going  2^  miles  an  hour  I  can  reach  my  destination 
in  12  hours.  How  long  would  I  be  in  making  the  trip  if  I 
traveled  3-J-  miles  an  hour?  A7is.  9  hours. 

47.  I  buy  40  gallons  of  cider  at  20  cents  a  gallon.  If  I 
had  got  it  for  5J  cents  a  gallon  cheaper,  how  many  more 
gallons  could  I  have  purchased  ?  Ans. .  IS^^j^  gallons. 

48.  I  bought  4  acres  of  land  at  $80  an  acre.  I  sold  A  20 
rods  square  and  B  20  square  rods  at  $150  per  acre.  How 
much  had  I  left,  and  what  were  my  receipts  for  the  part 
sold  ?  Ans.  Had  left  220  sq.  rods.     Received  $393f. 

49.  If  I  gain  ^  ct.  apiece  by  selling  eggs  at  14  cts.  a  doz. , 
how  much  will  I  gain  on  each  egg  if  I  sell  them  at  20  cts. 
a  doz.  ?  A71S.  1  ct. 

50.  If  I  sell  oats  at  42|-  cts.  per  bushel,  my  gain  is  only  f 
of  what  it  would  be  if  I  should  sell  them  at  56:^  cts.  per 
bushel.     What  did  they  cost  me  ?  Ans.  15^  per  bu. 

398.  51.  On  Monday  pig  iron  was  selling  at  $23  per  ton. 
Tuesday  it  advanced  20^  ;  and  the  next  day  15^  on  Tues- 
day's price.     What  was  the  price  on  Wednesday  ? 

Ans.  $31.74. 

52.  A  merchant  shipped  1350  bushels  of  wheat  in  March, 
18^  more  in  April,  and  10^  more  in  May  than  in  April. 
How  many  bushels  did  he  ship  during  the  three  months 
named  ?  A7is.  4695.3  bu. 

53.  Mr.  Murray's  oil-well  yielded  as  follows  for  four  suc- 
cessive days  :  1st  day,  500  bbls. ;  2d  day,  700  bbls. ;  3d  day, 
9%  more  than  on  the  two  preceding  days ;  and  on  the  4th  day, 
15^  more  than  on  the  three  preceding  days.  How  many 
bbls.  did  the  well  yield  in  the  four  days  ?       Ans.  5392.2. 

54.  Four  men  owned  a  boat  worth  $25000.  The  part 
owned  by  the  first  was  60^  of  the  value  of. the  boat;  the 
part  owned  by  the  second  was  equal  to  40;^  of  that  owned 


MISCELLANEOUS      PROBLEMS.  435 

by  the  first ;  the  part  owned  by  the  third  was  equal  to  16-|^ 
of  that  owned  by  the  second  ;  and  the  part  o^wied  by  the 
fourth  was  equal  to  300^  of  that  owned  by  the  third.  What 
was  the  value  of  each  man's  share  ? 

Arts,  1st,  115000 ;  2d,  $6000  ;  3d,  $1000  ;  4th,  $3000. 
300.  55.  In  1878  Mr.  Ferguson  traveled  2750  miles ; 
in  '79  he  traveled  550  miles  more  than  in  '78 ;  in  '80  he 
traveled  2120  miles  more  than  in  '79,  and  in  '81  3260  miles 
less  than  in  '80.  Find  how  many  %  more  or  less  he  traveled 
in  '79,  '80,  and  "81  than  in  '78. 

20^  more  in  '79  ;    97^\%  more  in  '80  ; 
^  ^^^*  21^^  less  in '81. 

56.  The  number  of  bushels  of  wheat  in  New  York  on 
March  19  was  1407000  bu.,  and  the  number  on  the  26th  of 
March  of  the  same  year  was  1063000  bu.  How  many  % 
less  bu.  were  in  New  York  on  the  latter  than  on  the  former 
date?  Ans.  24.45  —  ^. 

57.  In  Chicago,  January  6  th,  1880,  in  the  morning  wheat 
was  selling  at  96^  per  bu. ;  in  the  afternoon  of  the  same  day 
it  was  selling  at  97 J^'  per  bu.     What  was  the  rate  %  increase? 

Ans.  1^%. 
30^.   58.  A  gentleman  who  invested  in  4^  U.  S.  Bonds 
at  par  has  a  yearly  income  from  these  bonds  of  $640.     What 
is  the  par  value  of  his  bonds  ?  A7is,  $16000. 

59.  Having  sold  half  a  section  of  land  to  one  purchaser, 
f  of  a  section  to  another,  ;^  of  }  of  a  section  to  another,  I 
find  that  I  have  only  15^  of  my  land  left.  How  many  acres 
had  I  at  first  ?  Ans.  988^  acres. 

60.  Having  paid  out  ^  of  my  salary  for  rent,  -J-  for  cloth- 
ing for  my  family,  5^  to  the  church,  20^  for  a  horse  and 
buggy,  I  have  $920  left.     What  is  my  salary  ? 

Ans.  $2400. 


436  misoella:s^eous    problems. 

61.  Mr.  Sampson  sold  his  watch  for  113  7-|-,  which  was  at  a 
discount  of*8-J-^.     At  what  price  did  he  value  his  watch  ? 

A?is.  $150. 

62.  Having  disposed  of  f  of  my  land,  I  donated  -|  of  the 
remainder  to  a  church,  75%  of  what  I  had  left  I  used  for  a 
building-lot  and  lawn,  and  then  had  left  28  perches  for  a 
cow-pasture.    How  many  acres  had  I  at  first  ? 

Ans.  6  A.  64  P. 

63.  Sold  a  sewing  machine  at  an  advance  on  cost  of  $20, 
which  was  equal  to  J  of  200^  of  the  cost.     Find  the  cost. 

Ans.  $30. 
-    64.   Mercer  sold  maple  molasses  to  Gaston  for  20^  per  gal- 
lon advance.     Gaston  sold  the  same  at  an  advance  of  25^  per 
gal.,  and  thus  gained  25^.     What  did  the  molasses  cost  Mer- 
cer per  gallon  ?  Ans.  80^. 

304.  65.  Having  added  20^  to  the  number  of  trees  in 
my  orchard,  I  find  that  I  now  have  720  trees.  How  many 
had  I  before  adding  the  20^  ?  Ans.  600  trees. 

Q6.  A  gentleman  having  been  in  business  two  years,  added 
20%  to  his  stock  at  tlie  end  of  the  first  year  ;  and  to  this  15% 
of  the  original  stock  at  the  end  of  the  second  year,  when  he 
found  that  his  stock  amounted  to  $16956.  What  was  his 
original  stock  ?  Ans.  %12bQ0. 

67.  The  Pacific  Ocean  has  a  surface  of  65630000  square 
miles,  which  is  88\-fJf %  more  than  the  number  of  square 
'miles  in  the  surface  of  the  Atlantic  Ocean.  How  many 
square  miles  in  the  Atlantic  ?  Ans.  34780000  sq.  mi. 

i  68.  The  length  of  the  Mackenzie  River  is  2120  miles, 
.(which  is  314t%  less  than  the  length  of  the  Amazon.  What 
is  the  length  of  the  Amazon?  Ans.  3080  mi. 

313.  69.  A  merchant  purchased  $20000  worth  of  goods 
from  a  wholesale  dealer,  with  the  understanding  that  if  he 


MISCELLAKEOUS      PROBLEMS.  437 

sold  at  an  advance  of  over  20f^,  the  wholesale  dealer  was  to 
receive  15^  of  the  excess  over  20;^.  The  good-s  were  sold 
at  an  average  advance  of  27^.  How  much  of  the  profits 
should  the  merchant  pay  over  ?  Ans.  $210. 

70.  A  nurseryman  purchased  1250  apple-trees  at  12J^ 
apiece.  When  selling  them  he  found  that  20^  of  them  were 
dead.  He  sold  the  remainder,  however,  so  as  to  make  37j-^ 
profit  on  his  investment.  At  how  much  apiece  did  he  sell 
the  trees  ?  A7is.  2H^  apiece. 

71.  In  selling  out  my  stock  of  cloths  I  am  willing  to  take 
an  average  of  1 2%  less  than  my  selling  price.  I  find,  how- 
ever, that  I  can  sell  125  yards  cassimeres,  selling  price  $1.50 
per  yard,  for  $1.40  per  yard;  and  75  yards  broadcloth,  sell- 
ing price  $3.50  per  yard,  at  $3.25.  At  what  price  can  I 
afford  to  sell  150  yards  Tricots,  selling  price  82.50  per  yard, 
so  that  my  average  discount  will  be  12^  ? 

Ans.  $2.04|-  per  yard. 

72.  Mr.  Boyd  purchased  five  horses,  paying  $125  for  the 
first,  $140  for  the  second,  $135  for  the  third,  $180  for  the 
fourth,  and  $225  for  the  fifth.  He  kept  them  all  for  five 
days  at  an  expense  of  50^  each  per  day ;  and  then  sold  the 
first  horse  for  $150,  the  second  for  $125,  the  third  for  $210, 
the  fourth  for  $215.  The  fifth  horse  he  had  to  keep  15  days 
longer  at  an  expense  of  50^  per  day,  when  he  finally  sold 
him  for  enough  to  make  his  average  gain  20;:^  on  the  lot. 
Find  the  selling  price  of  the  fifth  horse.  Ans.  $290. 

215.  73.  A  drover  having  1520  sheep  valued  at  $2.75  per 
head,  exchanged  them  for  1930  sheep  which  he  sold  at  the 
rate  of  $2.75  per  head.  How  many  %  did  he  gain  by  the 
operation?  Ans.  2'7%  nearly. 

74.  In  January  apples  sold  for  $2.50  per  bbl. ;  in  February 


438  MISCELLANEOUS      PROBLEMS. 

at  13.00  per  bbl. ;   in  March  at  $3.35  per  bbl. ;  and  in  April 
at  15.25.    What  was  the  rate  %  advance  each  month  ? 

Ans.  20^,  llf^,  56ff^. 

75.  Smith  bought  a  farm  for  $5360.  He  sold  the  farm 
to  Moorhead  at  an  advance  of  1500.  Moorhead  sold  it  to 
Peebles  for  $9000.  Peebles  sold  it  to  Meredith  so  as  to  lose 
$450.  What  was  Smith's  and  Moorhead's  gain  %,  and  what 
was  Peebles'  loss  %  ?  (  Smith's  gain  9|f  ^  ; 

Ans.  I  Moorhead's  gain  53f|J^ ; 
(  Peebles'  loss  b%. 

76.  Mr.  Robb  purchased  150  acres  of  land  at  $60  per  acre. 
During  the  first  year  he  erected  on  his  land  a  dwelling  worth 
$2100,  a  barn  costing  $950,  and  corn-crib  and  other  out- 
buildings at  an  expense  of  $410.  He  paid  out  also  $375  for 
help,  taxes,  etc.  He  raised  150  bu.  potatoes,  which  brought 
35^  per  bu. ;  400  bu.  corn,  worth  40^  per  bu. ;  and  300  bu. 
wheat,  which  he  sold  at  85^  per  bu.  At  the  end  of  the  year 
he  sold  the  farm  for  $14292.75.  Leaving  out  of  the  ques- 
tion his  own  services  for  superintending,  etc.,  what  %  did  he 
gain?  A71S.  lb%. 

211.  77.  Mr.  Sands  sold  a  house  that  cost  him  $435  less 
than  his  selling  price,  at  an  advance  of  15^  on  cost.  What 
did  the  house  cost  him  ?  Aiis.  $2900. 

78.  Having  purchased  a  farm  of  175  acres,  I  added  in  the 
way  of  improvements  a  new  kitchen  to  the  house  at  a  cost 
of  $350  ;  a  new  carriage-house,  valued  at  $400  ;  190  rods  of 
board  fence  at  $1.60  per  rod  ;  and  a  picket  fence  around  my 
garden  at  a  cost  of  $210.  I  then  found  that  the  cost  of  my 
improvements  amounted  to  7J^  of  the  cost  of  my  farm. 
What  was  the  cost  of  the  farm,  and  how  much  has  it  now 
cost  me  per  acre  ?  Ans.  $16853^ ;  $103.52f}. 

79.  By  paying  cash  for  my  goods  I  can  purchase  at  a  dis- 


MISCELLANEOUS      PROBLEMS.  439 

count  of  0% ;  by  paying  in  30  days  I  can  obtain  a  reduction 
of  3%.  Having  purchased  a  bill  of  goods  I  paid  for  it  in  30 
days,  and  found  that  I  had  to  pay  $56.36  more  than  if  I  had 
paid  cash.     What  was  the  amount  paid  by  me  ? 

Aus.  $2733.46. 

80.  A  merchant  retiring  from  business  sold  out  his  stock 
and  fixtures  at  a  discount  of  31^  of  their  original  cost,  and 
realized  from  the  sale  $3156.     What  was  their  original  cost? 

Ans.  $4573.91  +  . 

219.  81.  Sold  in  one  year  dry  goods  to  the  amount  of 
$36125,  and  gained  on  an  average  15^.  What  was  the  cost 
of  the  goods  ?  Ans.  $31413.04/^. 

82.  Captain  Martin  having  purchased  a  steamboat,  had  it 
fitted  up  at  an  expense  of  $3150.25,  and  then  sold  it  for 
$21344,  thereby  making  a  gain  of  16^.  What  did  the  cap- 
tain pay  for  the  boat  ?  Ans.  $15249.75. 

83.  Henry  Macomb  purchased  a  piece  of  coal-land,  for 
which  he  paid  with  borrowed  money.  The  use  of  the 
money  cost  $175.20,  and  he  sold  the  land  for  $30572.15, 
making  a  gain  of  13^  on  his  entire  investment.  What  was 
the  first  cost  of  the  land  ?  Ans.  $26879.80. 


84.  Sold  I  of  a  rolling-mill  for  $166763.50,  and  gained 
18;^.     What  was  the  value  of  the  entire  mill  ? 

Ans.  $226120. 

85.  What  price  must  I  ask  for  my  horse,  which  cost  me 
$130,  that  I  may  fall-in  price  12^^^  and  yet  make  20^? 

Ans.  $178f 

86.  Sold  my  house  and  lot  for  $5640,  which  was  20^ 
more  than  its  cost.  At  what  price  should  I  have  sold  it  to 
make  25^  on  cost?  Ans.  $5875. 

87.  A  man  sold  his  saw  and  grist-mill  for  $9922,  which 


440  MISCELLANEOUS      PROBLEMS. 

was  1S%  below  cost.     How  many  %  would  he  have  gained 
had  he  sold  it  for  $14217.50  ?  Ajis.  l?!^. 

88.  A  manufacturer  sells  broadcloth  to  a  merchant  at  an 
advance  of  15^  on  cost;  the  merchant  sells  to  the  tailor  at 
20^  increase  of  manufacturers'  prices ;  and  the  tailor  sells 
to  his  customer  at  an  advance  of  33^%  on  the  merchant's 
price.  At  what  price  does  the  manufacturer  make  goods 
that  cost  the  tailor's  customer  $3.68  per  yard? 

Ans.  $2.00. 

89.  Sold  a  span  of  horses  and  a  carriage  for  $620.  Had 
I  sold  them  for  $662.50,  I  would  have  gained  8^%  more. 
How  many  %  did  I  gain  ?  Ans.  24^. 

90.  Sold  a  town  lot  at  a  loss  of  12^^.  Had  it  cost  me 
$102  less,  I  would  have  gained  10^.  What  was  the  cost  of 
the  lot?  A71S.  $498f. 

91.  Sent  a  cargo  of  Texas  beef  to  Liverpool,  Eng.,  and 
sold  it  at  a  profit  of  29^.  Invested  the  proceeds  in  railroad 
iron,  which  I  sold  in  New  York  for  $48709.11,  at  a  profit  of 
22^.    What  was  the  cost  of  the  cargo  of  beef  ? 

Ans.  $30950. 

92.  Sold  two  houses,  receiving  $3750  for  each.  On  one 
I  gained  20;^,  and  on  the  other  lost  20^.  How  much  greater 
was  my  loss  than  my  gain  ?  Ans.  $312.50. 

93.  A  gentleman  having  purchased  a  number  of  Kentucky 
horses,  sold  them  so  as  to  gain  18|^  on  his  investment. 
With  the  proceeds  he  purchased  grain,  which  he  sold  at  an 
advance  of  12^% ;  and  gained  by  the  two  operations  $860. 
What  was  the  amount  invested  in  Kentucky  horses  ? 

Ans.  $2560. 

94.  An  oil  merchant  shipped  petroleum  from  Ncav  York 
to  Liverpool.  The  oil  cost  him  9J^  per  gal.  Freight  and 
other  charges  amounted  to  2f  ^  per  gal.     He  sold  the  oil  at 


MISCELLANEOUS      PROBLEMS.  441 

25^  per  imperial  gal.  (=  1.2  U.  S.  gal.).     What  %  was  his 
profit?  Ans.  731^^. 

95.  When  refined  petroleum  is  selling  in  New  York  for 
7|^  per  gal.,  and  in  London  at  6|d.  per  imperial  gal.,  what 
%  is  left  for  profit  after  'dZ\%  is  allowed  for  freight,  charges, 
insurance,  etc.  ?  Ans.  4.6%'. 

96.  Keceived  from  Mrs.  Adolph  a  consignment  of  oranges, 
embracing  350  boxes.  Of  these  I  sold  to  Perkins  75  boxes, 
at  $2.50  per  box;  to  Jenkins  142  boxes,  at  $2.75  per  box; 
and  the  remainder  to  Smallman  at  $2.87J  per  box.  After 
taking  out  my  percentage  at  ^^%,  and  $60  for  freight  and 
charges,  how  much  should  I  remit  to  Mrs.  Adolph  ? 

Ans.  $866.76  +  . 

97.  Purchased  for  Mr.  Blair  1000  bu.  wheat  at  95^  per  bu., 
2120  bu.  corn  at  45^  per  bu.,  3500  bu.  oats  at  31^  per  bu., 
and  paid  for  sacks,  storage,  drajage,  etc.  $750.  What 
amount  should  Mr.  Blair  remit  to  me  to  pay  for  the  grain, 
my  commission  at  ^%,  and  expenses  ? 

Ans.  $3828.67. 

98.  Jas  McDonald  having  sent  me  1327  bbls.  mess  pork 
to  sell  on  commission,  I  charged  for  various  expenses  $250  ; 
and  after  deducting  my  commission,  I  remitted  to  him 
$21745.02-J.  Having  sold  the  pork  at  $17  per  bbl.,  what 
rate  %  commission  did  I  charge  him  ?  Ans.  2^%. 

99.  I  sold  for  Mr.  McFarland  580  bu.  potatoes  at  75^  per 
bu.,  1048  bbls.  apples  at  $3.25  per  bbl.,  and  590  bu.  corn  at 
70^  pel  bu  After  taking  out  my  commission  and  $125  for 
charges,  I  sent  $3982.04  to  Mr.  McFarland.  W4iat  %  com- 
mission did  I  charge  him  ?  Ans.  3^%. 

100.  An  insurance  agent,  after  paying  $545  for  office  rent, 
$135  for  advertising,  and  $925  for  commissions  to  sub-agents, 
found  that  he  had  cleared  during  the  year  $3741.     What 


442  MISCELLA]S^EOUS      PEOBLEMS. 

was  the  amount  of  his  collections,  his  commission  being 
15^  ?  Ans.  $35640. 

101.  My  agent  in  Liverpool  having  sold  for  me  a  cargo  of 
grain,  sent  me  a  bill  for  $2653,  of  which  amount  $1375  were 
for  freight  and  other  expenses,  and  the  balance  his  commis- 
sion at  6%.    At  what  price  did  he  sell  the  grain  ? 

Ans.  $21300. 

102.  A  real  estate  agent  charged  me  $301.50  for  renting 
my  houses  and  collecting  my  rents,  his  commission  being 
6%  ;  for  selling  some  Western  land  at  7%  commission  his  bill 
Avas  128.49 ;  for  purchasing  a  store  in  the  city,  his  charges  at 
21%  commission  were  $25.50.  How  much  rent  did  he  col- 
lect ?  what  was  the  value  of  the  Western  land  ?  and  what 
price  was  paid  for  the  store  ?    Ans.  $6030 ;  $407 ;  $1020. 

103.  My  agent  in  Baltimore  having  sold  for  me  a  con- 
signment of  grain,  after  paying  a  bill  of  $1125  for  freight 
and  taking  out  his  commission  at  3%,  sent  me  a  draft  for 
$19536.     At  what  price  was  the  grain  sold  ? 

Ans.  $21300. 

104.  A  broker  was  instructed  to  sell  $4000  worth  of  U.  S. 
4:%  bonds  at  par,  and  invest  the  proceeds  in  P.  R.  R.  stock 
at  $50  a  share.  He  charged  f%  for  selling  the  bonds,  i%  for 
investing  the  proceeds,  and  $.25  for  postage.  How  many 
shares  did  he  purchase  ?  Ans.  79  shares. 

105.  Sent  from  Chicago  to  New  York  1930  bu.  of  wheat, 
which  my  factor  sold  at  $1.30  per  bu.  After  paying  $62 
charges  and  deducting  his  commission  at  2%,  he  invested  the 
remainder  in  coffee  at  16^  per  lb.,  his  commission  on  the 
purchase  of  coffee  being  1^%.  How  many  lbs.  of  coffee  did 
he  purchase  ?  Ans.  15140.5 -f- lb. 

106.  A  real  estate  dealer  in  Washington  sold  my  mansion 
for  $78500,  on  which  he  charged  his  commission  of  3^^; 


MISCELLANEOUS      PKOBLEMS.  443 

and  sent  me  my  money  by  a  draft,  which  I  had  discounted 
at  ^%.     What  did  I  reahze  from  the  sale  of  my  house  ? 

A?is.  $75468.43  +  . 

107.  Shipped  to  Liverpool  a  cargo  of  cotton,  which  was 
sold  by  my  correspondent  for  £12500,  and  the  money  invested 
in  broadcloths.  The  commission  charged  for  selling  the  cot- 
ton was  3%,  and  for  purchasing  the  cloth  2^%.  The  freight 
of  both  cotton  and  cloths  was  £700,  and  was  paid  out  of  the 
proceeds  of  the  sale  of  the  cotton.  What  was  the  value  in 
U.  S.  money  of  the  cloth  ?  Atis.  $54243.67-+-. 

108.  Sold  land  for  James  Kenan,  charging  him  4:%  com- 
mission, and  invested  proceeds  in  bank  stocks  at  1^%  com- 
mission. My  commissions  amounted  to  $297.  At  what 
price  did  I  sell  the  land  ?  Ans.  $5481. 

109.  Mr.  Hanson  sold  my  span  of  horses,  charging  me  6% 
commission,  and  invested  the  proceeds  in  groceries,  com- 
mission 5^.  His  two  commissions  amounted  to  $50.  How 
much  did  he  invest  in  groceries?  Ans.  $525. 

110.  A  broker  during  one  year  charged  his  customers  for 
selling  bank  stocks,  etc.,  |^ ;  and  for  investing  the  proceeds 
of  these  sales  in  IJ.  S.  bonds  he  charged  f  ^.  His  commis- 
sions of  this  kind  amounted  to  $650.  What  was  the  value 
of  stocks,  etc.,  sold  by  him  ?  Ans.  $40300. 

111.  Mr.  Knox  sent  me  a  bill  of  $10703.26,  which  included 
a  mortgage  which  he  had  purchased  for  me,  with  his  com- 
mission at  21%  ;  $50  for  examining  title,  $2.50  for  recording, 
and  $1. 75  for  telegrams  und  postage.  What  was  the  face  of 
the  mortgage  ?  Ans.  $10364. 

112.  My  agent  charged  me  3^%  for  selling  my  iron,  and 
S%  for  investing  the  proceeds  in  wool.  His  bill  for  com- 
missions amounted  to  $292.50.  At  what  price  did  he  sell 
the  iron?  Ans.  $46e35. 


444 


MISCELLANEOUS      PROBLEMS. 


353.  Find  the  interest  of 


Principal. 

For 

At 

Interest, 

113. 

116740 

5yr. 

9  mo. 

Q% 

$5775.30. 

114. 

$1250.53 

2yr. 

11  mo. 

6% 

$218.84  +  . 

115. 

110540. 

10  mo. 

6% 

$527.00. 

116. 

11360 

lyr. 

1  mo.  15  da. 

6% 

$91.80. 

117. 

$1360 

4  yr. 

1  mo.  15  da. 

4^ 

$224.40. 

118. 

S232.71 

3  yr. 

7  mo. 

4^ 

$33.355  4-. 

119. 

$930.84 

3  yr. 

7  mo. 

^% 

$166.78-. 

120. 

6971.10 

lyr. 

5  mo.  17  da. 

^% 

$510.25—. 

121. 

$652 

1  mo.    8  da. 

1% 

$4,816. 

122. 

$960.18 

2yr. 

4  mo. 

n 

$156,828. 

123. 

$7218 

lyr. 

2  mo.    6  da. 

n 

$600,376. 

124. 

$432.10 

5  yr. 

4  mo.  24  da. 

H% 

$81.66  +  . 

125. 

$16740 

5  yr. 

9  mo. 

H% 

$4331.475. 

126. 

$794.32 

lyr. 

6  mo.    1  da. 

n% 

$131.30  +  . 

127. 

$7645.76 

lyr. 

7  mo.  13  da. 

Qi% 

$804.82  +  . 

128. 

$3924.55 

6yr. 

1  mo.    5  da. 

9^ 

$2153.60  +  . 

129. 

$1728.19 

lyr. 

5  mo.  10  da. 

12% 

$299,552. 

130. 

$397.16 

lyr. 

6  mo.    1  da. 

H% 

$32.82  +  . 

131. 

$1250.53 

lyr. 

5  mo.  15  da. 

H% 

$113.98. 

132. 

$6375.06 

2yr. 

3  mo.    1  da. 

H% 

$969.41-. 

133. 

$23127.84 

6  mo.  25  da. 

n% 

$954.83-. 

134. 

$8919.07 

lyr. 

8  mo.    6  da. 

n% 

$1126.03  +  . 

135. 

$7248 

2yr. 

4  mo.  12  da. 

H% 

$600,376. 

136. 

$590.90 

2yr. 

5  mo.    5  da. 

d% 

M3.09-. 

137. 

$150 

lyr. 

4  mo.    6  da. 

H% 

$7.09-. 

138. 

$200 

2yr. 

6  mo.     6  da. 

H% 

^$22.65. 

139. 

$847.15 

3yr. 

3  mo.    2  da. 

4^ 

$110.32-. 

140. 

$792.30 

lyr. 

1  mo.    9  da. 

H% 

$39.51  +  . 

MISCELLANEOUS      PROBLEMS.  445 

Find  the  exact  interest  of 


Principal. 

$7500.00 

From 

To 

At 

Interest. 

141. 

May    1,  '76 

Apr.     3,  '77 

^% 

$415.48. 

142. 

$11242.00 

Jan.    1,  '77 

July     3,  '77 

Vo 

$394.55. 

143. 

$1077.50 

Jan.    8,  '75 

May     8,  '77 

7.3^ 

$183. 17i. 

144. 

$1854.80 

May  4,  '71 

June    9,  '77 

4^ 

$452,467. 

145. 

$3350.00 

July   5,  '76 

June  10,  '77 

^% 

$171.63. 

146. 

$2835.50 

Mcli.  6,  '71 

May   18,  '72 

H% 

$153,117. 

147.  On  the  2d  of  December,  1877,  I  borrowed  from  my 
friend,  Mr.  James,  $627.50.  On  the  5th  of  August,  1879,  I 
paid  him  $701.03,  which  included  the  money  borrowed,  with 
interest  for  the  time  the  money  was  in  my  possession.  What 
rate  ^  of  exact  interest  did  I  pay  ?  Ans.  7^. 

148.  On  June  7th,  1879,  Mr.  Sloan  borrowed  $2160.30; 
and  on  Aug.  19,  1880,  he  paid  principal  and  exact  interest 
amounting  to  $2328.80.  What  rate  %  did  he  pay  for  the  use 
of  the  money  ?  Ans.  6^%. 

149.  H.  Rogers  paid  for  the  use  of  a  certain  amount  of 
money  for  1  yr.  1  mo.  15  da.,  at  6%,  $45.90.  What  was  the 
amount  used?  Ans.  $680. 

150.  James  Frey  borrowed  on  Jan.  1,  1880,  at  8%,  a  sum 
of  money,  for  which  he  paid  on  May  22,  1881,  principal  and 
interest,  $2092.64.     What  was  the  sum  borrowed  ? 

Ans.  $1883. 

151.  John  Marsh  had  the  use  of  a  sum  of  money  from 
Jan.  31,  1875,  to  March  3,  1876,  at  6%  interest.  On  the 
latter  date  he  paid  principal  and  interest  amounting  to 
$929.45.    What  sum  of  money  did  he  have  the  use  of  ? 

Ans.  $872.45. 


446  MISCELLANEOUS      PROBLEMS. 

361.  152.  Having  borrowed  $7248,  for  which  I  was  to 
pay  1%  interest,  I  returned  to  the  lender  at  the  end  of  the 
period  for  which  I  had  borrowed,  principal  and  interest 
amounting  to  18448.752.  For  what  time  had  I  the  use  of 
the  money?  Ans.  2  yr.  4  mo.  12  da. 

153.  Having  borrowed  1680  from  a  friend,  I  retained  the 
money  until  the  interest  at  4^  was  $112.20.  How  long  did 
I  retain  the  money  ?  Ans.  4  yr.  1  mo.  15  da. 

154.  A  gentleman  borrowed  11675  on  July  5,  1877,  for 
which  he  was  to  pay  b^%  interest.  At  the  time  of  payment 
principal  and  exact  interest  amounted  to  11760.815.  On 
what  date  was  payment  made?  Ans.  June  10,  1878. 

374.  155.  $2500yOoV  Boston,  March  21,  1879. 
Three  months  after  date  I  promise  to  pay  to  the  order  of 

James  Wickline  $2500,  with  interest  at  6^,  value  received. 

Thos.  Hartman. 

On  this  note  were  the  following  indorsements  :  June  24, 1879,  $300  ; 
Aug.  17,  1879,  $400  ;   Dec.  6,  1879,  $500  ;  June  3,  1880,  $16 ;   Dec.  30, 

1880,  $500. 

What  was  due  Feb.  1,  1881?  Ans.  $970.38  +  . 

375.  156.  $18001^0-        Harrisbueg,  March  5,  1880. 
One  day  after  date  I  promise  to  pay  to  Samuel  Jenkins, 

or  order.  Eighteen  Hundred  ^f^  Dollars,  with  interest  from 
date,  without  defalcation,  value  received. 

Simon  Partridge. 

On  this  note  were  the  following  indorsements:  May  11,  1880,  $300  ; 
June  5,  1880,  $100  ;  July  17,  1880,  $200 ;  August  23, 1880,  $150  ;  Sept. 
29,  1880,  $75 ;  Oct.  17,  1880,  $200  ;  Dec.  29,  1880,  $300  ;  Feb.  17,  1881, 
$400. 

What  was  due  March  5,  1881,  one  year  from  the  time  the 

note  was  made  ?  Ans.  $240.85. 

378.   157.  A  gentleman  invested  in  a  savings  bank  for 


MISCELLAN^EOUS      PROBLEMS.  447 

his  son's  benefit  $1500,  on  his  son's  15th  birthday.  This 
sum  remained  on  compound  interest  at  b%  till  the  son  was 
23  years  old.     To  what  did  it  then  amount  ? 

Ans,  12216. 18  +  . 

158.  Having  deposited  in  a  savings  bank  which  pays  3^ 
semi-annually,  to  what  amount  am  I  entitled  at  the  end  of 
40  years,  provided  I  deposited  1300,  and  allowed  both  prin- 
cipal and  interest  to  remain  undisturbed  for  the  entire 
period?  Ans.  $3192.25  +  . 

394.  159.  I  sold  to  the  Citizens'  National  Bank  my  note 
having  3  mo.  to  run  at  a  discount  of  Q%.  The  face  of  the 
note  was  $425.     What  did  I  receive  for  the  note  ? 

Ans,  $418.41. 

160.  $7500-jfg-.  Pittsburgh,  Jan.  3,  1881. 
Four  months  after  date  I  promise  to  pay  to  Samuel  Magee, 

or  order.  Seven  thousand  five  hundred  dollars,  without  de- 
falcation, value  received.  Michael  Maginist. 

The  above  note  was  discounted  at  the  Bank  of  Pittsburgh 
on  March  1,  1881,  at  Q%  discount.  How  much  was  the  dis- 
count ?  Ans.  $82.50. 

161.  $3500.  READiiyTG,  Apr.  5,  1880. 
One  year  after  date  I  promise  to  pay  to  the  order  of 

Thomas  Hamilton,  Three  thousand  five  hundred  dollars 
with  interest,  without  defalcation,  for  value  received. 

Hiram  Mansfield. 
This  note  was  discounted  Jan.  3,  1881,  at  7^.     What  were 
the  proceeds  ?  Ans.  $3643.19  +  . 

162.  $3400,  New  York,  May  1,  1880. 
Nine  months  after  date  I  promise  to  pay  to  the  order  of 

Wm.  Simms  $3400  with  interest,  value  received. 

James  Martin. 


448  MISCELLANEOUS      PROBLEMS. 

The  above  note  was  discounted  at  the  Metropolitan  Bank, 
Nov.  9,  1880,  at  ^%  (exact)  discount.  What  were  the  pro- 
ceeds ?  Ans.  $3519.96  +  . 

396.  163.  A  friend  discounted  for  me  a  note  calling  for 
1960.18  in  14  months,  and  charged  me  $78.97+  discount. 
What  rate  %  did  he  charge  ?  A71S.  1%. 

164.  Paid  the  Bank  of  Commerce  $72.18—  for  discount- 
ing a  note  calling  for  $3250  in  four  months  from  the  date  of 
discount.     What  rate  %  did  the  bank  charge  ?     Ans.  6^%. 

398.  165.  Mr.  Burns  having  a  note  due  in  three  months, 
sold  it  to  the  Third  National  Bank  at  7%  discount,  and 
received  as  proceeds  $805.26.  What  was  the  face  of  the 
note  ?  Ans.  $820.09—. 

166.  James  Morrison  had  a  note  discounted  at  bank,  at 
8%  (exact)  discount.  The  note  was  dated  Jan.  1,  1881,  and 
read,  "  Three  months  after  date  I  promise,"  etc.  The  pro- 
ceeds were  $1004.11  + .     What  was  the  face  of  the  note  ? 

Ans.   $1025. 

400.  167.  For  what  sum  must  I  make  a  note,  due  in 
4  mo.  from  May  6,  1881,  in  order  to  realize  by  its  sale  at  7% 
(exact)  discount  $1094.51  ?  Ans.  $1121.67. 

168.  Having  purchased  a  bill  of  goods  amounting  to 
$1824.42,  I  pay  for  the  same  by  a  note  dated  Jan.  1,  '81,  and 
due  in  three  months,  which,  when  discounted  at  once  at  7% 
(exact),  will  pay  the  bill  in  full.  What  must  be  the  face  of 
this  note  ?  Ans.  $1857.55. 

402.  169.  On  the  4th  of  May,  1880,  I  sold  a  note  whose 
face  was  $820.15  and  dated  May  4,  1880,  at  7%  discount,  and 
received  for  it  $805.     When  was  this  note  nominally  due  ? 

Ans.  Aug.  4,  1880. 

170.  March  15,  1881,  I  purchased  from  a  neighbor  a  note 


MISCELLAlfTEOUS       PKOBLEMS.  449 

for  $919.94  at  an  (exact)  discount  of  1%,  and  gave  him  for 
it  $899.30.     When  Avas  the  note  legally  due  ? 

Ans.  July  10,  1881. 

410.  171.  I  have  a  mortgage  for  $2500,  bearing  6^  simple 
interest.  This  mortgage  has  two  years  to  run,  but  I  am 
willing  to  sell  it  now  at  a  discount  of  ^%.  What,  then,  do 
I  regard  as  the  present  worth  of  the  mortgage  ? 

A71S.  $2413.79  +  . 

172.  Sold  my  interest  in  a  woolen  mill  for  $5000,  to  be 
paid  for  in  three  years  from  date  of  sale,  with  simple  interest 
at  b%.  The  mortgage  securing  payment  of  the  $5000  with 
interest  I  sold  to  Mr.  Bradley  at  such  a  price  that  he  could 
realize  on  his  investment  %%  compound  interest.  What  did 
I  receive  for  the  mortgage  ?  Ans.  $4827.81  H- . 

413.  173.  What  cost  54  shares  W.  E.  &  Co.'s  Express 
stock,  at  18|-;^  premium,  brokerage  \%,  par  value  $100  per 
share?  Aris,  $6432.75. 

174.  When  U.  S.  4's  are  selling  at  113|;  what  cost  bonds 
having  a  face  of  $5500,  when  the  brokerage  is  1%  ? 

Ans.  $6276J. 

175.  Purchased  75  shares  Chicago  and  Alton  bonds  for 
$10350,  par  value  $100.  What  %  premium  did  I  pay  for 
them  ?  Ans.  3S%. 

176.  Sold  Central  Pacific  bonds  representing  $3000  for 
$3442.50.     What  rate  %  premium  did  I  obtain  ? 

Ans.  Ui%. 

177.  Having  purchased  55  shares  N.  Y.  Central  at  145, 
brokerage  |^,  I  sold  them  150,  brokerage  Y/o-  ^^^  value 
being  $100  per  share,  how  much  did  I  make  by  the  opera- 
tion ?  Ans.  $213. 12^. 

178.  At  what  price  must  I  sell  stocks  representing  a 
par  value  of  $4500,  brokerage  for  purchasing  ^%,  and  for 


450  MISCELLAI^EOUS      PROBLEMS. 

selling  1%,  $2.25  for  postage  and  telegrams,  so  as  to  clear 
1750?  '  Atis.  $5325. 37f 

179.  Sent  my  broker  $27827.50,  with  instructions  to  invest 
in  P.  F.  W.  and  0.  R.  R.  bonds.  He  informed  me  that  he 
had  to  pay  a  premium  of  32^  on  the  bonds  ;  and  that  his 
charges  were  $2.50  for  postage,  telegrams,  etc.;  and  his 
brokerage  i%.     How  many  bonds  did  he  purchase  for  me  ? 

Ans.  21  $1000  bonds. 

180.  When  U.  S.  4's  are  selling  at  a  premium  of  13  J;^,  and 
1^%  niust  be  paid  for  brokerage,  how  many  $500  bonds  can 
be  purchased  for  $6331. 87J  ?  Ans.  11. 

181.  Sold  125  shares  Boston  and  Maine  R.  R.,  par  $100, 
at  $101,  brokerage  i%-;  and  invested  the  proceeds  in  P.  R.  R. 
stock  at  62^,  brokerage  f  %'.  How  many  shares  of  P.  R.  R. 
stock  (par  $50)  did  I  purchase?  Ans.  200  shares. 

182.  What  rate  %  do  I  make  when  I  pay  $120  for  $100 
bonds  which  bring  8%  of  their  face  ?  Ans.  6^%. 

183.  When  U.  S.  4's  are  selling  at  112|,  what  rate  of 
interest  do  they  pay  the  investor  ?  A7is.  S^%. 

184.  Which  is  better— to  invest  in  U.  S.  5's  at  120  or 
U.  S.  3's  at  96  ?     Ans.  The  former  in  the  ratio  of  4^  to  3^. 

185.  At  what  price  should  a  4^  bond  be  sold,  that  it 
would  yield  the  same  interest  as  a  3^  bond  at  par  ? 

186.  I  sold  my  U.  S.  5's  at  an  advance  of  18^,  and  in- 
vested the  proceeds  in  bank  stocks  yielding  '7%  dividends. 
How  many  ^  did  I  increase  ray  income  from  the  money  in- 
vested in  U.  S.  5*s  ?  Ans.   65^^. 

187.  Sold  my  telegraph  stock  (par  100)  paying  10^  divi- 
dends at  150.  I  invested  the  proceeds  in  insurance  stocks 
(par  100)  paying  8%  dividends,  so  as  to  increase  my  income 


MISCELLANEOUS      PROBLEMS.  451 

^%.      How  much  per  share  did  I  pay  for  the  insurance 
stock?  Ans.  $114f. 

188.  Sold  my  bank  stock  (par  100),  which  paid  4^  divi- 
dends, at  a  discount.  Invested  the  proceeds  in  bonds  bear- 
ing ^%  interest,  which  I  purchased  at  a  premium  of  10^. 
I  thus  increased  my  income  4^.  At  what  rate  %  discount 
did  I  sell  my  bank  stock?  Ans.  42.8^. 

427.  189.  What  costs  a  30  days'  draft  on  New  York  for 
$3600,  at  1%  premium,  interest  off  at  b%  ?       Ans,  $3603. 

190.  What  costs  in  New  York  a  30  days  draft  on  Liver- 
pool for  £800,  at  f ^  premium,  interest  off  at  6^,'  estimating 
£1  =  M. 8665  ?  Ans.  $3900. 986  + . 

191.  What  costs  a  60  days  draft  in  Bremen  for  6000  marks, 
exchange  Y/o  discount,  and  interest  off  at  6^  ? 

(1  mark  =  33.8  cents.)  Ans.  $1404.081. 

192.  If  it  costs  me  $4715.72+  for  a  60  days  draft  on 
Olasgow  for  £975  12s.,  interest  off  at  Q%,  what  rate  %  pre- 
mium do  I  pay?  A71S.   %%. 

193.  Wishing  to  send  a  remittance  of  $5670  to  Cawnpore, 
India,  I  purchased  from  a  banker  a  4  months  draft,  paying 
Mm  1^%  for  exchange,  and  he  allowed  me  Q%  interest. 
What  was  the  face  of  the  draft  in  rupees,  each  equal  to  43. 6 
eents  ?  Ans.  12933+  rupees. 

445.  194.  The  town  of  Mohawk  containing  1250  taxables, 
real  estate  assessed  at  $5376475,  personal  property  amounting 
to  $937540,  propose  to  raise  for  general  expenses,  $3540 ; 
for  school  purposes,  $1675 ;  and  for  new  town-hall,  $3000. 
After  levying  a  poll-tax  of  $1.25  per  head,  how  many  % 
must  be  levied  on  the  personal  property  and  real  estate,  3% 
being  allowed  for  lost  taxes  and  b%  for  collecting  ? 

Ans.  1.164+  mills  on  the  $1. 


452  MISCELLANEOUS      PKOBLEMS. 

468.  195.  Maynard  &  Co.  made  an  assignment  in  favor  of 
their  creditors.  It  was  found  that  their  liabilities  amounted 
to  $74600,  and  their  assets  were  10000  bbls.  crude  oil,  worth 
$1.12^  per  bbl.;  three  oil-barges,  worth  $750  each  ;  1  steam- 
boat, sold  for  $10500;  oil  territory,  worth  $6500;  and 
accounts,  notes,  etc.,  which  brought  at  sale  $7560.  Ex- 
penses of  advertising,  etc.,  amounted  to  $1716|,and  assignee's 
charges  were  b%.  What  %  did  the  creditors  receive,  and  how 
much  did  Mr.  Nobbs  obtain,  whose  claim  was  $3000  ? 

Ans.  4:6i%;  Mr.  Nobbs  received  $1385. 

510.  19Q.  A  gentleman  travelling  abroad  saw  some  fine  car-^ 
pet,  and  concluded  to  purchase  enough  to  cover  the  floor  of 
a  square  room  that  he  had  at  home ;  but  the  only  dimension 
he  could  remember  was  the  diagonal  of  the  room,  which 
was  21  ft.  8  in.     How  many  yards  were  in  the  floor  ? 

Ans.  26  yd.+. 

197.  The  gable  end  of  my  house  is  16^ft.  wide,  the  "  comb  "" 
of  the  roof  is  6  ft.  above  the  "square."  How  long  must 
my  rafters  be  to  allow  2  ft.  for  the  width  of  the  eaves  ? 

Ans.   12  ft. 

198.  The  distance  from  my  right  eye  to  the  tip  of  the 
middle  finger  of  my  right  hand  is  2  ft.  9  in.  Now,  if  I  hold 
in  my  hand  a  pencil,  I  find  that  1  inch  of  the  pencil  held 
vertically  just  hides  from  my  view  a  tree  which  I  know  to  he 
31ft.  high.    Howfar  am  Ifrom  thetree?      Ans.  1023  ft. 

514.  199.  I  increased  my  "  hay-mow  "  by  doubling  the- 
amount  of  material  used  for  sides,  floors,  and  ceiling,  each 
retaining  its  shape.     How  much  did  I  increase  its  capacity? 

A?is.  2.83  times,  nearly. 


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